solve each equation by using the quadratic · 2018-09-17 · solve using the quadratic formula....

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Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary. 1. x 2 2x 15 = 0 SOLUTION: For this equation, a = 1, b = 2, and c = 15. The solutions are 5 and 3. 2. x 2 10x + 16 = 0 SOLUTION: For this equation, a = 1, b = 10, and c = 16. The solutions are 8 and 2. 3. x 2 8x = 10 SOLUTION: Write the equation in standard form. For this equation, a = 1, b = 8, and c = 10. The solutions are 6.4 and 1.6. 4. x 2 + 3x = 12 SOLUTION: Write the equation in standard form. For this equation, a = 1, b = 3, and c = 12. The solutions are 2.3 and 5.3. 5. 10x 2 31x + 15 = 0 SOLUTION: For this equation, a = 10, b = 31, and c = 15. The solutions are 2.5 and 0.6. 6. 5x 2 + 5 = 13x SOLUTION: Write the equation in standard form. For this equation, a = 5, b = 13, and c = 5. The solutions are 0.5 and 2.1. Solve each equation. State which method you eSolutions Manual - Powered by Cognero Page 1 2 - 5 Solving Quadratic Equations by Using the Quadratic Formula

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Page 1: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

eSolutions Manual - Powered by Cognero Page 1

2-5 Solving Quadratic Equations by Using the Quadratic Formula

Page 2: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

eSolutions Manual - Powered by Cognero Page 2

2-5 Solving Quadratic Equations by Using the Quadratic Formula

Page 3: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

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2-5 Solving Quadratic Equations by Using the Quadratic Formula

Page 4: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

eSolutions Manual - Powered by Cognero Page 4

2-5 Solving Quadratic Equations by Using the Quadratic Formula

Page 5: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

eSolutions Manual - Powered by Cognero Page 5

2-5 Solving Quadratic Equations by Using the Quadratic Formula

Page 6: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

eSolutions Manual - Powered by Cognero Page 6

2-5 Solving Quadratic Equations by Using the Quadratic Formula

Page 7: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

eSolutions Manual - Powered by Cognero Page 7

2-5 Solving Quadratic Equations by Using the Quadratic Formula

Page 8: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

eSolutions Manual - Powered by Cognero Page 8

2-5 Solving Quadratic Equations by Using the Quadratic Formula

Page 9: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

eSolutions Manual - Powered by Cognero Page 9

2-5 Solving Quadratic Equations by Using the Quadratic Formula

Page 10: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

eSolutions Manual - Powered by Cognero Page 10

2-5 Solving Quadratic Equations by Using the Quadratic Formula

Page 11: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

eSolutions Manual - Powered by Cognero Page 11

2-5 Solving Quadratic Equations by Using the Quadratic Formula

Page 12: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

eSolutions Manual - Powered by Cognero Page 12

2-5 Solving Quadratic Equations by Using the Quadratic Formula

Page 13: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

eSolutions Manual - Powered by Cognero Page 13

2-5 Solving Quadratic Equations by Using the Quadratic Formula

Page 14: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

eSolutions Manual - Powered by Cognero Page 14

2-5 Solving Quadratic Equations by Using the Quadratic Formula

Page 15: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

eSolutions Manual - Powered by Cognero Page 15

2-5 Solving Quadratic Equations by Using the Quadratic Formula

Page 16: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

eSolutions Manual - Powered by Cognero Page 16

2-5 Solving Quadratic Equations by Using the Quadratic Formula

Page 17: Solve each equation by using the Quadratic · 2018-09-17 · Solve using the quadratic formula. Write the equation in standard form. For this equation, a = 2, b = ±8, and c = ±12

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

1. x2 − 2x − 15 = 0

SOLUTION:  For this equation, a = 1, b = –2, and c = –15.

The solutions are 5 and –3.

2. x2 − 10x + 16 = 0

SOLUTION:  For this equation, a = 1, b = –10, and c = 16.

The solutions are 8 and 2.

3. x2 − 8x = −10

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –8, and c = 10.

The solutions are 6.4 and 1.6.

4. x2 + 3x = 12

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = 3, and c = –12.

The solutions are 2.3 and –5.3.

5. 10x2 − 31x + 15 = 0

SOLUTION:  For this equation, a = 10, b = –31, and c = 15.

The solutions are 2.5 and 0.6.

6. 5x2 + 5 = −13x

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 13, and c = 5.

The solutions are –0.5 and –2.1.

Solve each equation. State which method you used.

7. 2x2 + 11x − 6 = 0

SOLUTION:  Solve by factoring: In this trinomial, a = 2, b = 11, and c = –6, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the  factors of 2(–6) or –12 and identify the factors with a sum of 11.

The correct factors are –1 and 12.   So, use m = -1 and p = 12.

Solve the equation using the Zero Product Property

The solutions are –6 and .

Factors of –12 Sum  1, –12 –11 –1, 12 11 2, –6 –4 –2, 6 4 3, –4 –1 –3, 4 1

8. 2x2 − 3x − 6 = 0

SOLUTION:  Solve using the quadratic formula. For this equation, a = 2, b = –3, and c = –6.Quadratic Formula:

The solutions are 2.6 and –1.1.

9. 9x2 = 25

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factoring:

The solutions are .

10. x2 − 9x = −19

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 1, b = –9, and c = 19.Quadratic Formula:

The solutions are 5.6 and 3.4.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

11. x2 − 9x + 21 = 0

SOLUTION:  For this equation, a = 1, b = –9, and c = 21.

The discriminant is –3. Since the discriminant is negative, the equation has no real solutions.

12. 2x2 − 11x + 10 = 0

SOLUTION:  For this equation, a = 2, b = –11, and c = 10.

The discriminant is 41. Since the discriminant is positive, the equation has two real solutions.

13. 9x2 + 24x = −16

SOLUTION:  Write the equation in standard form.

For this equation, a = 9, b = 24, and c = 16.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

14. 3x2 − x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = –1, and c = –8.

The discriminant is 97. Since the discriminant is positive, the equation has two real solutions.

15. TRAMPOLINE Eva springs from a trampoline to dunk a basketball. Her height h in feet can be

modeled by the equation h = –16t2 + 22.3t + 2,

where t is time in seconds. Use the discriminant to determine if Eva will reach a height of 10 feet. Explain.

SOLUTION:  Write the equation in standard form.

For this equation, a = –16, b = 22.3, and c = –8.

  The discriminant is −14.91. Since the discriminant is negative, the equation has no real solutions. Thus, Eva will not reach a height of10 feet.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

16. 4x2 + 5x − 6 = 0

SOLUTION:  For this equation, a = 4, b = 5, and c = –6.

The solutions are  and –2.

17. x2 + 16 = 0

SOLUTION:  For this equation, a = 1, b = 0, and c = 16.

The discriminant is negative, so the equation has no real solutions, ø.

18. 6x2 − 12x + 1 = 0

SOLUTION:  For this equation, a = 6, b = –12, and c = 1.

The solutions are 1.9 and 0.1.

19. 5x2 − 8x = 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = –8, and c = –6.

The solutions are 2.2 and –0.6.

20. 2x2 − 5x = −7

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –5, and c = 7.

The discriminant is negative, so the equation has no real solutions, Ø.

21. 5x2 + 21x = −18

SOLUTION:  Write the equation in standard form.

For this equation, a = 5, b = 21, and c = 18.

The solutions are  and –3.

22. 81x2 = 9

SOLUTION:  Write the equation in standard form.

For this equation, a = 81, b = 0, and c = –9.

The solutions are .

23. 8x2 + 12x = 8

SOLUTION:  Write the equation in standard form.

For this equation, a = 8, b = 12, and c = –8.

The solutions are 0.5 and –2.

24. 4x2 = −16x − 16

SOLUTION:  Write the equation in standard form.

For this equation, a = 4, b = 16, and c = 16.

The solution is –2.

25. 10x2 = −7x + 6

SOLUTION:  Write the equation in standard form.

For this equation, a = 10, b = 7, and c = –6.

The solutions are 0.5 and –1.2.

26. −3x2 = 8x − 12

SOLUTION:  Write the equation in standard form.

For this equation, a = –3, b = –8, and c = 12.

The solutions are –3.7 and 1.1.

27. 2x2 = 12x − 18

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = –12, and c = 18.

The solution is 3.

28. AMUSEMENT PARKS The Demon Drop at Cedar Point in Ohio takes riders to the top of a towerand drops them 60 feet. A function that approximates

this ride is h = −16t2 + 64t − 60, where h is the height

in feet and t is the time in seconds. About how many seconds does it take for riders to drop 60 feet?

SOLUTION:  

−16t2 + 64t − 60 = 0

For this equation, a = –16, b = 64, and c = –60.

It takes 2.5 seconds for the riders to drop from 60 feet to 0 feet.

Solve each equation. State which method you used.

29. 2x2 − 8x = 12

SOLUTION:  Solve using the quadratic formula. Write the equation in standard form.

For this equation, a = 2, b = –8, and c = –12.Use the Quadratic Formula.

The solutions are 5.2 and –1.2.

30. 3x2 − 24x = −36

SOLUTION:  Solve by using the quadratic formula. Write the equation in standard form.

For this equation, a = 3, b = –24, and c = 36.Use the Quadratic Formula.

The solutions are 6 and 2.

31. x2 − 3x = 10

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solutions are 5 and –2.

32. 4x2 + 100 = 0

SOLUTION:  Solve using the quadratic formula. First, check the value of the discriminant. For this equation, a = 4, b = 0, and c = 100.

The discriminant is negative, so the equation has no real solutions, ø.

33. x2 = −7x − 5

SOLUTION:  Solve by completing the square.

First, isolate the x- and the x2-terms.

  The solutions are –0.8 and –6.2.

34. 12 − 12x = −3x2

SOLUTION:  Solve by factoring. Write the equation in standard form.

Factor.

The solution is 2.

State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.

35. 0.2x2 − 1.5x + 2.9 = 0

SOLUTION:  For this equation, a = 0.2, b = –1.5, and c = 2.9.

The discriminant is –0.07. Since the discriminant is negative, the equation has no real solutions.

36. 2x2 − 5x + 20 = 0

SOLUTION:  For this equation, a = 2, b = –5, and c = 20.

The discriminant is –135. Since the discriminant is negative, the equation has no real solutions.

37. x2 − x = 3

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = – , and c = –3.

The discriminant is 12.64. Since the discriminant is positive, the equation has two real solutions.

38. 0.5x2 − 2x = −2

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.5, b = –2, and c = 2.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

39. 2.25x2 − 3x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.25, b = –3, and c = 1.

The discriminant is 0. Since the discriminant is 0, the equation has one real solution.

40. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 2, b = , and c = .

The discriminant is 18.25. Since the discriminant is positive, the equation has two real solutions.

41. CCSS MODELING The percent of U.S. households with high-speed Internet h can be

estimated by h = −0.2n2 + 7.2n + 1.5, where n is the

number of years since 1990. a. Use the Quadratic Formula to determine when 20% of the population will have high-speed Internet. b. Is a quadratic equation a good model for this information? Explain.

SOLUTION:  a. Write the equation in standard form.

For this equation, a = –0.2, b = 7.2, and c = –18.5.

Since n is the number of years since 1990 add the solutions to 1990.  1990 + 3 = 1993; 1990 + 33 = 2023 b.  A quadratic equation not a good model for this information. The parabola has a maximum at about 66, meaning only 66% of the population would ever have high-speed Internet.

[-3. 48] scl: 5, by [-2, 68] scl: 7

42. TRAFFIC The equation d = 0.05v2 + 1.1v models

the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. If Hannah’s car stopped after 250 feet on a highway with a speed limit of 65 miles per hour, was she speeding? Explain your reasoning.

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.05, b = 1.1, and c = –250.

No, she was not speeding; Sample answer: Hannah was traveling at about 61 mph, so she was not speeding.

Without graphing, determine the number of x-intercepts of the graph of the related function for each function.

43. 4.25x + 3 = −3x2

SOLUTION:  Write the equation in standard form.

For this equation, a = 3, b = 4.25, and c = 3.

The discriminant is –17.9. Since the discriminant is negative, the graph of the function will not have any x-intercepts.

44. 

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = , and c = .

The discriminant is 0.04. Since the discriminant is positive, the graph of the function will have two x-intercepts.

45. 0.25x2 + x = −1

SOLUTION:  Write the equation in standard form.

For this equation, a = 0.25, b = 1, and c = 1.

The discriminant is 0. Since the discriminant is 0, the graph of the function will have one x-intercept.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.

46. −2x2 − 7x = −1.5

SOLUTION:  Write the equation in standard form.

For this equation, a = –2, b = –7, and c = 1.5.

The solutions are –3.7 and 0.2.

47. 2.3x2 − 1.4x = 6.8

SOLUTION:  Write the equation in standard form.

For this equation, a = 2.3, b = –1.4, and c = –6.8.

The solutions are 2.1 and –1.4.

48. x2 − 2x = 5

SOLUTION:  Write the equation in standard form.

For this equation, a = 1, b = –2, and c = –5.

The solutions are 3.4 and –1.4.

49. POSTER Bartolo is making a poster for the dance. He wants to cover three fourths of the area with text.

a. Write an equation for the area of the section with text. b. Solve the equation by using the Quadratic Formula. c. What should be the margins of the poster?

SOLUTION:  a.  The length of the area covered by text 20 – (x + x) or 20 – 2x. The width of the area is 25 – (4x + 3x)

or 25 – 7x.  The area of the entire poster is 20 in. × 25 in., or 500 in

2. Because Bartolo wants the area

with text to be three-fourths of the total area, it must

be ft. So, an equation for the area of

the section with text is (20 − 2x)(25 − 7x) = 375. b. Write the equation in standard form.

For this equation, a = 14, b = –190, and c = 125.

The solutions are 12.9 and 0.7. c. The margins should be about 0.7 in. on the sides, 4 • 0.7 = 2.8 in. on the top, and 3 • 0.7 = 2.1 in. on the bottom.

50. MULTIPLE REPRESENTATIONS In this problem, you will investigate writing a quadratic equation with given roots. 

If p is a root of 0 = ax2 + bx + c, then (x – p ) is a

factor of ax2 + bx + c.

a. Tabular Copy and complete the first two columns of the table. b. Algebraic Multiply the factors to write each equation with integral coefficients. Use the equationsto complete the last column of the table. Write each equation. c. Analytical How could you write an equation with three roots? Test your conjecture by writing an equation with roots 1, 2, and 3. Is the equation quadratic? Explain.

SOLUTION:  a. For any two roots m and p , in the left hand column,the middle column will be the corresponding factors (x – m), (x – p ).    b. The equation with these factors will be: (x – m)(x

– p ) = 0 which simplifies to x2 – (m + p )x + mp = 0.

Use this to fill in the column of the table. 

c. You could write an equation with three roots by multiplying the corresponding factors together and setting it equal to zero. If an equation has the three roots 1, 2, 3, then the corresponding factors would be(x – 1), (x – 2), and (x – 3). The equation would thenbe:

  This is not a quadratic equation since it is of degree 3.     

51. CHALLENGE Find all values of k such that 2x2 −

3x + 5k = 0 has two solutions.

SOLUTION:  For the equation to have two solutions, the discriminant must be positive.

52. REASONING Use factoring techniques to

determine the number of real zeros of f (x) = x2 − 8x

+ 16. Compare this method to using the discriminant.

SOLUTION:  

For f(x) = x2 − 8x + 16, a = 1, b =  −8 and c = 16.

Then the discriminate is b2 − 4ac or (−8)

2−4(1)(16)

= 0.  The polynomial can be factored to get f (x) = (x

− 4)2.

Solve to find the real zeros.

So the only real zero is 4. The discriminant is 0, so the only real zero is 4. The discriminant is 0, so there is 1 real zero. The discriminant tells us how many real zeros there are. Factoring tells us what they are.

CCSS STRUCTURE  Determine whether there are two, one, or no real solutions.

53. The graph of a quadratic function does not have an x-intercept.

SOLUTION:  If there are no x-intercepts, then there are no real solutions.

54. The graph of a quadratic function touches but does not cross the x-axis.

SOLUTION:  If the graph is tangent to the x-axis, meaning there is only one x-intercept, then there is only one real solution.

55. The graph of a quadratic function intersects the x-axis twice.

SOLUTION:  If there are two x-intercepts, then there are two real solutions.

56. Both a and b are greater than 0 and c is less than 0 in a quadratic equation.

SOLUTION:  

The discrimininant is b2 – 4ac. No matter the value

of b, b2 will always be positive. If  a is greater than 0

and c is less than 0, then – 4ac will be positive. Thus the discrimininant would be positive. So there would be two real solutions.

57. WRITING IN MATH Why can the discriminant beused to confirm the number of real solutions of a quadratic equation?    

SOLUTION:  

Consider a few examples: ,

.   For the first equation we have a = 1, b = 1, and c = 6.  

  In this case the discriminant is negative and the roots are imaginary.    For the second equation we have: a = 1, b = 1, and c

= –6.

  When the discriminant is positive, the roots are real. We can only have imaginary roots if we take the square root of a negative number, which will only happen if the discriminant is negative. 

58. WRITING IN MATH Describe the advantages and disadvantages of each method of solving quadratic equations. Which method do you prefer, and why?

SOLUTION:  Factoring: Factoring is easy if the polynomial is factorable and complicated if it is not. Not all equations are factorable. 

For example f (x) = x2 – 8x + 16 factors to (x – 4)

2.

However, f (x) = x2 – 16x + 8 can not be factored.

  Graphing: Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x-term. 

For example, for the quadratic f (x) = 2x2 – 17x + 4,

you can see the two solutions in the graph. However,it will be difficult to identify the solution x = 8.2578049 in the graph. 

[-5, 15] scl: 2 by [-30, 10] scl: 4   Completing the square: Completing the square can be used for any quadratic equation and exact solutions can be found, but the

leading coefficient has to be 1 and the x2 - and x-

term must be isolated. It is also easier if the coefficient of the x-term is even; if not, the calculations become harder when dealing with

fractions.  For example x2 + 4x = 7 can be solved by

completing the square. 

  Quadratic Formula: The Quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.  For example, use the 

Quadratic Formula to find the solutions of f (x) = 4x2

+ 13 x + 5. 

  See students’ preferences.

59. If n is an even integer, which expression represents the product of three consecutive even integers? A n(n + 1)(n + 2) B (n + 1)(n + 2)(n + 3) C 3n + 2 D n(n + 2)(n + 4)

SOLUTION:  Adding 2 to an even integer results in the next consecutive integer. n n + 2

n + 2 + 2, or n + 4 So, the correct choice is D.

60. SHORT RESPONSE The triangle shown is an isosceles triangle. What is the value of x?

SOLUTION:  Because an isosceles triangle has two equal angles, xcould be equal to 64, or it could be equal to the unnamed angle, where 180 = 2x + 64.

The value of x is 58 or 64.

61. Which statement best describes the graph of x = 5? F It is parallel to the x-axis. G It is parallel to the y-axis. H It passes through the point (2, 5). J It has a y-intercept of 5.

SOLUTION:  A line for which x is always 5, is a vertical line, which is parallel to the y-axis. So, the correct choice is G.

62. What are the solutions of the quadratic equation 6h2

+ 6h = 72? A 3 or −4 B −3 or 4 C no solution D 12 or −48

SOLUTION:  Write the equation in standard form.

For this equation, a = 6, b = 6, and c = –72.

The solutions are –4 or 3. So, the correct choicer is A.

Solve each equation by completing the square. Round to the nearest tenth if necessary.

63. 6x2 − 17x + 12 = 0

SOLUTION:  

The solutions are  or  .

64. x2 − 9x = −12

SOLUTION:  

The solutions are 1.6 or 7.4.

65. 4x2 = 20x − 25

SOLUTION:  

Describe the transformations needed to obtain the graph of g (x) from the graph of f (x).

66. f (x) = 4x2

g(x) = 2x2

SOLUTION:  

The graph of g(x) = ax2 stretches or compresses the

graph of f (x) = 4x2 vertically. The change in a is ,

and 0 <  < 1. If 0 <   < 1, the graph of f (x) = x2

is compressed vertically. Therefore, the graph of y =

2x2 is the graph of y = 4x

2 vertically compressed.

67. f (x) = x2 + 5

g(x) = x2 − 1

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is –6, and –6 < 0. If c < 0, the graph of f

(x) = x2 is translated  units down. Therefore, the 

graph of y = x2 – 1 is a translation of the graph of y

= x2 +5 shifted down 6 units.

68. f (x) = x2 − 6

g(x) = x2 + 3

SOLUTION:  

The graph of f (x) = x2 + c represents a vertical

translation of the parent graph. The value of the change in c is 9, and 9 > 0. If c > 0, the graph of f (x)

= x2 is translated  units up. Therefore, the graph of

y = x2 +3 is a translation of the graph of y = x

2 –6

shifted up 9 units.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

69. 20, 25, 30, ...

SOLUTION:  Check the difference and ratio between terms.    25 – 20 = 5 30 – 25 = 5   There is a common difference of 5 between the terms. The sequence is arithmetic.

70. 1000, 950, 900, ...

SOLUTION:  Check the difference and ratio between terms.    950 – 1000 = –50 900 – 950 = –50    There is a common difference of –50 between the terms. The sequence is arithmetic.

71. 200, 350, 650, …

SOLUTION:  Check the difference and ratio between terms.    350 – 200 = 150 650 – 350 = 200   There is no common difference between the terms.   

  There is no common ratio. This is neither an arithmetic nor geometric sequence.     

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2-5 Solving Quadratic Equations by Using the Quadratic Formula