lesson 4.8 , for use with pages 292-299

Lesson 4.8, For use with pages 292-299

1. Write 15x2 + 6x = 14x2 – 12 in standard form.

2. Evaluate b2 – 4ac when a = 3, b = –6, and c = 5.

ANSWER

ANSWER –24

x2 + 6x +12 = 0

3. A student is solving an equation by completing thesquare. Write the step in the solution that appearsjust before “(x – 3) = 5.”

ANSWER (x – 3)2 = 25

Lesson 4.8, For use with pages 292-299

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Questions 4.7

EXAMPLE 1 Solve an equation with two real solutions

Solve x2 + 3x = 2.

x2 + 3x = 2 Write original equation.

x2 + 3x – 2 = 0 Write in standard form.

x = – b + b2 – 4ac2a

Quadratic formula

x = – 3 + 32 – 4(1)(–2)

2(1)a = 1, b = 3, c = –2

Simplify.x = – 3 + 17

2

The solutions are x = – 3 + 172

0.56 and

x = – 3 – 17

2– 3.56.

ANSWER

EXAMPLE 1 Solve an equation with two real solutions

CHECK

Graph y = x2 + 3x – 2 and note that the x-intercepts are about 0.56 and about – 3.56.

EXAMPLE 2 Solve an equation with one real solutions

Solve 25x2 – 18x = 12x – 9.

25x2 – 18x = 12x – 9. Write original equation.

Write in standard form.

x = 30 + (–30)2 – 4(25)(9)

2(25)a = 25, b = –30, c = 9

Simplify.

25x2 – 30x + 9 = 0.

x =30 + 0

50

x =35 Simplify.

35

The solution is

ANSWER

EXAMPLE 2 Solve an equation with one real solutions

CHECK

Graph y = –5x2 – 30x + 9 and note that the only x-intercept is 0.6 = . 3

5

EXAMPLE 3 Solve an equation with imaginary solutions

Solve –x2 + 4x = 5.

–x2 + 4x = 5 Write original equation.


x = – 4+ 42– 4(– 1)(– 5)

2(– 1)a = –1, b = 4, c = –5

Simplify.

–x2 + 4x – 5 = 0.

x =– 4+ – 4

– 2

– 4+ 2ix =

– 2

Simplify.

Rewrite using the imaginary unit i.

x = 2 + i

The solution is 2 + i and 2 – i.

ANSWER

EXAMPLE 3 Solve an equation with imaginary solutions

CHECK

Graph y = 2x2 + 4x – 5. There are no x-intercepts. So, the original equation has no real solutions. The algebraic check for the imaginary solution 2 + i is shown.

–(2 + i)2 + 4(2 + i) = 5?

–3 – 4i + 8 + 4i = 5?

5 = 5

GUIDED PRACTICE for Examples 1, 2, and 3

Use the quadratic formula to solve the equation.

x2 = 6x – 41.

Write original equation.


x = – b + b2 – 4ac2a

Quadratic formula

x = – (– 6) + (– 6)2 – 4(1)(4) 2(1)

a = 1, b = – 6, c = 4

Simplify.x = + 3 + 20

2

x2 – 6x + 4 = 0

x2 = 6x – 4

SOLUTION


ANSWER

The solutions are x = 3 + 20

2 and

x = 3 – 20

2= 3 – 5

= 3 + 5



4x2 – 10x = 2x – 92.



x = – b + b2 – 4ac2a

Quadratic formula

x = – (– 12) + (– 12)2 – 4(4)(9) 2(4)

a = 4, b = – 12, c = 9

Simplify.x = 12 + 0

8

4x2 – 10x = 2x – 9

4x2 – 12x + 9 = 0

SOLUTION


ANSWER

32

The solution is = 12

1



7x – 5x2 – 4 = 2x + 33.



x = – b + b2 – 4ac2a

Quadratic formula

x = – (5) + (5)2 – 4(– 5)(–7)

2(– 5)a = – 5, b = 5, c = – 7

Simplify.–115x = – 5 +

– 10

7x – 5x2 – 4 = 2x + 3

– 5x2 + 5x – 7 = 0

SOLUTION


Simplify.

Rewrite using the imaginary unit i.

115x = – 5 + i

– 10

115x = 5 + i

10

ANSWER

The solutions are and . 115 5 + i

10

115 5 – i

10

EXAMPLE 4 Solve a quadratic inequality using a table

Solve x2 + x ≤ 6 using a table.

SOLUTION

Rewrite the inequality as x2 + x – 6 ≤ 0. Then make a table of values.

Notice that x2 + x –6 ≤ 0 when the values of x are between –3 and 2, inclusive.

The solution of the inequality is –3 ≤ x ≤ 2.

ANSWER

EXAMPLE 5 Solve a quadratic inequality by graphing

Solve 2x2 + x – 4 ≥ 0 by graphing.

SOLUTION

The solution consists of the x-values for which the graph of y = 2x2 + x – 4 lies on or above the x-axis. Find the graph’s x-intercepts by letting y = 0 and using the quadratic formula to solve for x.

0 = 2x2 + x – 4

x = – 1+ 12– 4(2)(– 4)

2(2)

x =– 1+ 33

4

x 1.19 or x –1.69

EXAMPLE 5 Solve a quadratic inequality by graphing

Sketch a parabola that opens up and has 1.19 and –1.69 as x-intercepts. The graph lies on or above the x-axis to the left of (and including) x = – 1.69 and to the right of (and including) x = 1.19.

The solution of the inequality is approximately x ≤ – 1.69 or x ≥ 1.19.

ANSWER

GUIDED PRACTICE for Examples 4 and 5

Solve the inequality 2x2 + 2x ≤ 3 using a table and using a graph. 5.

SOLUTION

Rewrite the inequality as 2x2 + 2x – 3 ≤ 0. Then make a table of values.

The solution of the inequality is –1.8 ≤ x ≤ 0.82.

ANSWER

x -3 -2 -1.8 -1.5 -1 0 0.5 0.8 0.9

22 + 2x – 3 9 1 -0.1 -1.5 -3 -3 -1.5 -0.1 0.42

EXAMPLE 6 Use a quadratic inequality as a model

The number T of teams that have participated in a robot-building competition for high school students can be modeled by

Robotics

T(x) = 7.51x2 –16.4x + 35.0, 0 ≤ x ≤ 9

Where x is the number of years since 1992. For what years was the number of teams greater than 100?

EXAMPLE 6 Use a quadratic inequality as a model

T(x) > 100

7.51x2 – 16.4x + 35.0 > 100

7.51x2 – 16.4x – 65 > 0

Graph y = 7.51x2 – 16.4x – 65 on the domain 0 ≤ x ≤ 9. The graph’s x-intercept is about 4.2. The graph lies above the x-axis when 4.2 < x ≤ 9.

There were more than 100 teams participating in the years 1997–2001.

ANSWER

SOLUTION

You want to find the values of x for which:

EXAMPLE 7 Solve a quadratic inequality algebraically

Solve x2 – 2x > 15 algebraically.

SOLUTION

First, write and solve the equation obtained by replacing > with = .

x2 – 2x = 15

x2 – 2x – 15 = 0

(x + 3)(x – 5) = 0

x = – 3 or x = 5

Write equation that corresponds to original inequality.


Factor.

Zero product property

EXAMPLE 7 Solve a quadratic inequality algebraically

The numbers – 3 and 5 are the critical x-values of the inequality x2 – 2x > 15. Plot – 3 and 5 on a number line, using open dots because the values do not satisfy the inequality. The critical x-values partition the number line into three intervals. Test an x-value in each interval to see if it satisfies the inequality.

Test x = – 4: Test x = 1:

12 –2(1) 5 –1 >15

Test x = 6:

The solution is x < – 3 or x > 5.

ANSWER

(– 4)2 –2(– 4) = 24 >15 62 –2(6) = 24 >15


6. Robotics

Use the information in Example 6 to determine in what years at least 200 teams participated in the robot-building competition.

SOLUTION

You want to find the values of x for which:

T(x) > 200

7.51x2 – 16.4x + 35.0 > 200

7.51x2 – 16.4x – 165 > 0


Graph y = 7.51x2 – 16.4x – 165 on the domain 0 ≤ x ≤ 9.

There were more than 200 teams participating in the years 1998 – 2001.

ANSWER


7. Solve the inequality 2x2 – 7x = 4 algebraically.

SOLUTION

First, write and solve the equation obtained by replacing > with 5.

Write equation that corresponds to original inequality.


Factor.

Zero product property

2x2 – 7x = 4

2x2 – 7x – 4 = 0

(2x + 1)(x – 4) = 0

x = – 0.5 or x = 4


The numbers 4 and – 0.5 are the critical x-values of the inequality 2x2 – 7x > 4 . Plot 4 and – 0.5 on a number line, using open dots because the values do not satisfy the inequality. The critical x-values partition the number line into three intervals. Test an x-value in each interval to see if it satisfies the inequality.

Test x = – 3: Test x = 2: Test x = 5:

– 3

– 4

– 2

– 1

0

1 2

3

4

5

6

7

– 5

– 6

– 7

2 (– 3)2 – 7 (– 3) > 4 2 (5)2 – 7 (3) > 4 2 (2)2 – 7 (2) > 4

The solution is x < – 0.5 or x > 4.

ANSWER

lesson 4.8 , for use with pages 292-299

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