simplifying radicals (for help, go to lessons 8-3 and 10-3.) algebra 1 lesson 11-1 complete each...

Simplifying RadicalsSimplifying Radicals

(For help, go to Lessons 8-3 and 10-3.)

ALGEBRA 1 LESSON 11-1ALGEBRA 1 LESSON 11-1

Complete each equation.

1. a 3 = a2 • a 2. b 7 = b6 • b

3. c 6 = c3 • c 4. d 8 = d4 • d

Find the value of each expression.

5. 4 6. 169 7. 25

8. 49

11-1

Simplifying RadicalsSimplifying RadicalsALGEBRA 1 LESSON 11-1ALGEBRA 1 LESSON 11-1

1. a3 = a(2 + 1) = a2 • a1 2. b7 = b(6 + 1) = b6 • b1

3. c6 = c(3 + 3) = c3 • c3 4. d8 = d(4 + 4) = d4 • d4

5. 4 = 2 6. 169 = 13

7. 25 = 5 8. 49 = 7

Solutions

11-1

Simplify 243.


11-1

243 = 81 • 3 81 is a perfect square and a factor of 243.

= 81 • 3 Use the Multiplication Property of Square Roots.

= 9 3 Simplify 81.

Simplify 28x7.


28x7 = 4x6 • 7x 4x6 is a perfect square and a factor of 28x7.

= 4x6 • 7x Use the Multiplication Property of Square Roots.

= 2x3 7x Simplify 4x6.

11-1


Simplify each radical expression.


a. 12 • 32 12 • 32 = 12 • 32 Use the Multiplication Property of

Square Roots.

= 384 Simplify under the radical.

= 64 • 6 64 is a perfect square and a factor of 384.

= 64 • 6 Use the Multiplication Property of

Square Roots.

= 8 6 Simplify 64.

11-1


(continued)


b. 7 5x • 3 8x

= 42x 10 Simplify.

= 21 • 2x 10 Simplify 4x2.

= 21 4x2 • 10 Use the Multiplication Property of

Square Roots.

= 21 4x2 • 10 4x2 is a perfect square and a

factor of 40x2.

7 5x • 3 8x = 21 40x2 Multiply the whole numbers and

use the Multiplication Property of

Square Roots.

11-1


Suppose you are looking out a fourth floor window 54 ft above

the ground. Use the formula d = 1.5h to estimate the distance you

can see to the horizon.


d = 1.5h

The distance you can see is 9 miles.

= 9 Simplify 81.

= 81 Multiply.

= 1.5 • 54 Substitute 54 for h.

11-1




= Simplify 64. 13

8

a. 1364

b. 49x4

7

x2 = Simplify 49 and x4.

11-1

= Use the Division Property of Square Roots.1364

13

64

= Use the Division Property of Square Roots.49x4

49

x4


= 12 Divide.120 10

= 4 • 3 4 is a perfect square and a factor of 12.

a. 120 10


= 4 • 3 Use the Multiplication Property of Square Roots.

= 2 3 Simplify 4.

11-1

b. 75x5

48x


= Divide the numerator and denominator by 3x.75x5

48x25x4

16

= Use the Division Property of Square Roots.25x4

16

(continued)

= Use the Multiplication Property ofSquare Roots.

25 • x4

16

= Simplify 25, x4, and 16.5x2

4

11-1

3

7

3

7

7

7 7

7= • Multiply by to make the denominator a

perfect square.



a. 3 7

= Simplify 49.3 7 7

11-1

= Use the Multiplication Property of Square Roots.3 7

49

= Simplify 36x4. 33x

6x2


(continued)

b. 11

12x3

11-1

Simplify the radical expression.

= • Multiply by to make the denominator a

perfect square.

3x

3x

3x

3x

11

12x3

11

12x3

= Use the Multiplication Property of Square Roots. 33x

36x4

26. 12 mi

27. 17 mi

28.

29.

30.

31.

32.

33.

34.


pages 581–583 Exercises

1. 10 2

2. 7 2

3. 5 3

4. –4 5

5. –6 30

6. 40 5

7. 2n 7

8. 6b2 3

9. 6x 3

10. 2n n

11. 2a2 5a

21

73 3

252

2 30

11

5

3a

7

4c

11-1

12. –4b2 3

13. 20

14. 18

15. 11 2

16. 84 3

17. 7 3

18. –30 3

19. 6n 2

20. 14t 2

21. 3x2 17

22. 80t 3

23. 3a3 2

24. –6a2 2

25. 3 mi

5 3a

7

45. 5

46.

47.

48.

49. 2 3

50.

51.

52. not simplest form; radical in the denominator of a fraction

53. not simplest form; radical in the denominator of a fraction

54. Simplest form; radicand has no perfect-square factors other than 1.

55. Simplest form; radicand has no perfect-square factors other than 1.

56. a. 18 • 10 = 180 = 36 • 5 = 6 5

b. Answers may vary. Sample: a = 36, b = 5; a = 9, b = 20

57. 30

58. 2 13

59.

2 10n

5n

35.

36. 3

37.

38. 2 3

39. –2 5

40. 2x 7

41.

42.

43.

44.


32

s3

a2 3

23y

3 2

2

21x

7x

9 2

4

2b

b

55y

2y

3 2

4

11-1

2n 2n

9


60.

61. 2 15

62.

63. – 3

64. 4 5

65. 2ab 5b

66. ab2c abc

67.

68.

–2 a

a2

x y

y 2

3m

4m8 6a

3a

77. 30a4

78. seconds

79. C

80. F

81. B

82. I

83. [2] A = 96 ft2 s = 96 = 16 • 6 = 4 6 ft

[1] correct answer, without work shown

84. quadratic; y = 0.2x2

85. exponential; y = 4(2.5)x

86. linear; y = –4.2x + 7

2 ± 10

3

11-1

69. –3 ± 3 2

70. 1 ± 5

71.

72. a. 50 = 25 • 2 = 25 • 2 = 5 2

b. The radicand has no perfect-square factors other than 1.

73. Answers may vary. Sample: 12, 27, 48.

74. a. 2 6 in.b. 4.90 in.

75. 12x

76. 10b2


87.

88.

89.

90. 3n2 + 5n + 5

91. 3v2 – v – 9

92. 5t 3 + 8t 2 – 14t – 11

93. –3b2 – 23b – 21

11-1


12

36


1. 16 • 8 2. 4 144 3.

4. 5. 2

a5

3x

15x3

8 2 48 3

3

2 a a3

5 5x

11-1

Simplify each expression.

1. 52 + 62 2. 92 – 42 3. (3t)2 + (4t)2

Solve each equation.

4. c2 = 36 5. 24 + b2 = 49 6. a2 + 16 = 65

7. 12 + b2 = 32 8. 80 = c2 9. 100 = a2 + 52

(For help, go to Lesson 10-4.)


The Pythagorean TheoremThe Pythagorean Theorem

11-2

The Pythagorean TheoremThe Pythagorean TheoremALGEBRA 1 LESSON 11-2ALGEBRA 1 LESSON 11-2

11-2

Solutions

1. 52 + 62 = (5 • 5) + (6 • 6) = 25 + 36 = 612. 92 – 42 = (9 • 9) – (4 • 4) = 81 – 16 = 653. (3t)2 + (4t)2 = (32 • t2) + (42 • t2) = 9t2 + 16t2 = 25t 2

4. c2 = 36 c = ± 36 = ±65. 24 + b2 = 496. a2 + 16 = 65

b2 = 49 – 24 a2 = 65 – 16b2 = 25 a2 = 49b = ± 25 = ±5 a = ± 49 = ±7

7. 12 + b2 = 328. 80 = c2

b2 = 32 – 12 c = ± 80 = ± 16 • 5 = ± 16 • 5 = ±4 5

b2 = 20b = ± 20 = ± 4 • 5 = ± 4 • 5 = ± 2 5

9. 100 = a2 + 52

100 – 52 = a2

48 = a2

a = ± 48 = ± 16 • 3 = ± 16 • 3 = ± 4 3


What is the length of the hypotenuseof this triangle?


a2 + b2 = c2 Use the Pythagorean Theorem.

82 + 152 = c2 Substitute 8 for a and 15 for b.

64 + 225 = c2 Simplify.

289 = c2 Find the principal square root of each side.

17 = c Simplify.

The length of the hypotenuse is 17 m.

11-2


A toy fire truck is near a toy building on a table such that the

base of the ladder is 13 cm from the building. The ladder is extended 28

cm to the building. How high above the table is the top of the ladder?


Define: Let b = height (in cm) of the ladder from a point 9 cm above the table.

Relate: The triangle formed is a right triangle. Use the Pythagorean Theorem.

11-2


Write: a2 + b2 = c2

132 + b2 = 282 Substitute.169 + b2 = 784 Simplify.

b2 = 615 Subtract 169 from each side.

b2 = 615 Find the principal square root of each side.

b 24.8 Use a calculator and round to the nearest tenth.

(continued)

The height to the top of the ladder is 9 cm higher than 24.8 cm, so it is about 33.8 cm from the table.

11-2


Determine whether the given lengths are sides of a right

triangle.


a. 5 in., 5 in., and 7 in.

This triangle is not a right triangle.

b. 10 cm, 24 cm, and 26 cm

This triangle is a right triangle.

100 + 576 676 Simplify. 676 = 676

102 + 242 262 Determine whether a2 + b2 = c2, where c is the longest side.

52 + 52 72 Determine whether a2 + b2 = c2, where c is the longest side.

25 + 25 49 Simplify. 50 = 49/

11-2


If two forces pull at right angles to each other, the

resultant force is represented as the diagonal of a rectangle,

as shown in the diagram. The diagonal forms a right triangle

with two of the perpendicular sides of the rectangle.

For a 50–lb force and a 120–lb force, the resultant force is 130 lb. Are the forces pulling at right angles to each other?

16,900 = 16,900

The forces of 50 lb and 120 lb are pulling at right angles to each other.

502 + 1202 1302 Determine whether a2 + b2 = c2 where c is the greatest force.

2500 + 14,400 16,900

11-2

12. 0.6

13. 1.2 m

14. about 15.5 ft

15. about 5.8 km

16. yes

17. no

18. no

19. yes

20. no

21. yes

22. yes

23. no




1. 10

2. 25

3. 17

4. 26

5. 2.5

6. 1

7. 4

8. 5

9. 12

10. 7.1

11. 7.5

24. no

25. yes

26. 1.5

27. or 0.3

28. 3

29. 6

30. 2.6

31. 7.0

32. a. 6 5 ftb. 80.5 ft2

33. yes

34. no

11-2

415



35. yes

36. yes

37. 4.2 cm

38. 1000 lb

39. 559.9

40. 9.0

41. 9.7

42. a. These lengths could be 2 legs or one leg and the hypotenuse.

b. about 12.8 in. or 6 in.

43. a. 62 + 82 = 36 + 64 = 100 = 102

b. 5; 12; 7; 41c. Answers may vary. Sample: 10, 24, 26

44. a. 6.9 ftb. 89.2 ft2

c. 981 watts

45. 12.8 ft

46. a. Answers may vary. Sample: 5, 20, 5

b. 5 units2

47. a. 13.4 ftb. 17.0 ftc. 10.6 ftd. No; the hypotenuse d must

be longer than each leg.

48. An integer has 2 as a factor; the integer is even; if an integer is even, then it has 2 as a factor; true.

11-2



49. A figure is a square; the figure is a rectangle; if a figure is a rectangle then the figure is a square; false.

50. You are in Brazil; you are south of the equator; if you are south of the equator you are in Brazil; false.

51. An angle is a right angle; its measure is 90°; if the measure of an angle is 90°, then it is a right angle; true.

52. 52 units2

53. 6 in.

54. 10

55. 4 3

56. 5

57. n2 + (n + 1)2 = (n + 2)2; 3, 4, 5

58. a.

b. 74

59. a. a2 + 2ab + b2

b. c2

c.

d. (a + b)2 = c2 + 4 ab ; a2 + 2ab + b2 = 2ab + c2; a2 + b2 = c2

e. This equation is the same as the Pythagorean Theorem.

ab2

11-2

12



69. 2b2 10b

70.

71.

72. 3 and 4

73. 8 and 9

74. –8 and –7

75. 11 and 12

76. rational

77. irrational

78. irrational

79. rational

22x2

2 6v

v 4

11-2

––> ––>

63

60. D

61. H

62. B

63. C

64. A

65. [2] It is a right triangle. Substitute 17, the longest side, for c and substitute the other lengths for a and b in the Pythagorean Theorem82 + 152 172 64 + 225 289 289 = 289

[1] incorrect equation OR incorrect explanation

66. 4 3

67.

68. 5 2



80. 8x2 – 4x

81. 12a2 + 15a

82. 18t 3 – 6t 2

83. –10p4 + 26p3

84. 15b3 + 5b2 – 45b

85. –7v4 + 42v2 – 7v

11-2


1. Find the missing length 2. Find the missing lengthto the nearest tenth. to the nearest tenth.

3. A triangle has sides of lengths12 in., 14 in., and 16 in. Is thetriangle a right triangle?

4. A triangular flag is attached to a post. The bottom of the flag is 48 in.above the ground. How far from the ground is the top of the flag?

16.6 5.7

no

57 in.

11-2

Find the length of the hypotenuse with the given leg lengths.If necessary, round to the nearest tenth.

1. a = 3, b = 4 2. a = 2, b = 5

3. a = 3, b = 8 4. a = 7, b = 5

For each set of values, find the mean.

5. x1 = 6, x2 = 14 6. y1 = –4, y2 = 8

7. x1 = –5, x2 = –7 8. y1 = –10, y2 = –3

The Distance and Midpoint FormulasThe Distance and Midpoint Formulas



11-3

The Distance and Midpoint FormulasThe Distance and Midpoint FormulasALGEBRA 1 LESSON 11-3ALGEBRA 1 LESSON 11-3

Solutions

1. a2 + b2 = c2 2. a2 + b2 = c2

32 + 42 = c2 22 + 52 = c2

9 + 16 = c2 4 + 25 = c2

25 = c2 29 = c2

c = 25 = 5 c = 29 5.4The length of the hypotenuse is 5. The length of the hypotenuse is about 5.4.

3. a2 + b2= c2 4. a2 + b2 = c2

32 + 82= c2 72 + 52 = c2

9 + 64 = c2 49 + 25 = c2

73 = c2 74 = c2

c = 73 8.5 c = 74 8.6The length of the hypotenuse is The length of the hypotenuse isabout 8.5. about 8.6.

5. mean = = = 10

6. mean = = = 2

6 + 14 2

20 2

–4 + 8 2

42

7. mean = = = –6

8. mean = = = –6.5

–5 + (–7) 2–10 + (–3) 2

–12 2–13 2

11-3

d = (9 – 6)2 + [–4 – (–9)]2 Substitute (9, –4) for (x2, y2) and (6, –9) for (x1, y1).

Find the distance between F(6, –9) and G(9, –4).


d = ( x2 – x1)2 + (y2 – y1)2 Use the distance formula.

d = 32 + 52 Simplify within parentheses.

d = 34 Simplify to find the exact distance.

The distance between F and G is about 5.8 units.

d 5.8 Use a calculator. Round to the nearest tenth.

11-3

Find the exact lengths of each side of quadrilateral EFGH.

Then find the perimeter to the nearest tenth.


EF = [4 – (–1)]2 + (3 + 5)2

= 52 + (–2)2

= 25 + 4

= 29

FG = (3 – 4)2 + (–2 – 3)2

= (–1)2 + (–5)2

= 1 + 25

= 26

GH = |–2 – 3| = 5 EH = [–2 – (–1)]2 + (–2 – 5)2

= (–1)2 + (–7)2

= 1 + 49

= 50

11-3

The perimeter = 29 + 26 + 5 + 50 22.6 units.


Find the midpoint of CD.

, = , Substitute (–3, 7) for(x1, y1) and (5, 2) for (x2, y2).

7 + 2

2

x1+ x2

2

y1+ y2

2

(–3) + 5

2

= , Simplify each numerator.22

92

= 1, 4 Write as a mixed

number.

92

12

The midpoint of CD is M 1, 4 .12

11-3

A circle is drawn on a coordinate plane. The endpointsof the diameter are (–3, 5) and (4, –3). What are the coordinates of the center of the circle?


, = , Substitute (–3, 5) for(x1, y1) and (4, –3) for (x2, y2).

5 + (–3)

2

x1+ x2

2

y1+ y2

2

(–3) + 4

2

The center of the circle is at , 1 .12

= , = , 112

22

12

11-3

22. MN 5.1; NP 3.6; MP = 5

23. JK = 4; KL 2.2; LJ 3.6

24. TU 7.6; UV 16.5; VT 13.3

25. a. OR = 29, ST = 29

b. ;

c. yes

26. a. (20, 80), (–40, 30)

b. 78.1 ft

c. (–10, 55) or 55L10




1. 15

2. 14

3. 10

4. 11.3

5. 2.8

6. 8.1

7. 16

8. 21.6

9. (1, 6)

10. (–1, 12)

11. (0, 0)

12. (–2, 3)

13. 5, –5

14. – 2 , –1

15. (–4, 4)

16. 8 , –9

17. 10.6

18. 9.4

19. AB 4.1; BC 3.2; AC = 5

20. DE 6.1; EF 3.6; DF 5.7

21. RS 3.2; ST 5.7; RT 5.1

12

12

12

52

52

11-3



1 + 42

52

27. Answers may vary. Sample: Suppose you have (1, 1) and (4, –3). To find the distance, square the difference betweenthe x-coordinates. Square the difference between the y-coordinates. Find the sum and take the square root, so 9 + 16 = 5. To find the midpoint, add x-coordinates together and divide by 2. Repeat for y-coordinates. So,

, 1 – 32

= , –1

28. Check students’ work.29. a.

b. (0, – )

c. one half mile south of the substation

30. about 9.5 km apart

31. a. 38.1 mib. 20 mi, 21.2 mic. 15 min, 16 min

32. Yes; all sides are congruent.

12

11-3

39. a. y = – x + 1

b.

c. 3

40. Yes; the distance from each point to the center is 5.

41. 19.1

42. 5

43. 22.9

44. –5

45. 3

46. –8



43

95

, – 75

47. 11.7

48. 11.2

49. 10.2

50. 26

51. 3.5

52. 8

53. 4.1

54. –14, 14

55. –10, 10

56. no solution

57. –5, 5

58. no solution

11-3

33. a. R(–27, –5)b. PR = 13 3.6

RQ = 13 3.6

34. a. about 4.3 mib. about 17.4 mi

35. a. M(–0.5, 3); N(5.5, 3)b. They are equal.

36. (x, y)

37. They are opposites.

38. a. 5 unitsb. Answers may vary. Sample:

(5, 0), (0, 5), (–5, 0), (0, –5), (3, –4), (–3, 4), (–3, –4)

c. circle



59. – ,

60. k2 + 11k + 24

61. v2 + 2v – 35

62. 2p2 – 17p – 9

63. 8w4 + 19w2 + 11

64. 7t3 + 5t2 + 5t – 2

65. 6c3 – 27c2 + 33c + 24

23

23

11-3


7.2

7.1

9.5 units

(9, – 3)

(1 , 4)12

1. Find the distance between M(2, –1) and N(–4, 3) to the nearest tenth.

2. Find the distance between P(–2.5, 3.5) and R(–7.5, 8.5) to the nearest tenth.

3. Find the midpoint of AB, A(3, 6) and B(0, 2).

4. Find the midpoint of CD, C(6, –4) and D(12, –2).

5. Find the perimeter of triangle RST to the nearest tenth of a unit.

11-3

Operations with Radical ExpressionsOperations with Radical Expressions




1. 52 2. 200 3. 4 54 4. 125x2

Rationalize each denominator.

5. 6. 7. 3

11

5

8 15

2x

11-4

Operations with Radical ExpressionsOperations with Radical ExpressionsALGEBRA 1 LESSON 11-4ALGEBRA 1 LESSON 11-4

11-4

Solutions

5. = • = =

6. = • = = = = = =

7. = • = =

3

11

3

11

11

11

3 • 11 33

11 • 11 11 5

8

5

8

8

8

5 • 8

8 • 8

4 • 10 8

4 • 10 8

40 8

2 10 8

10 4

15

2x

15

2x

2x

2x

15 • 2x

2x • 2x

30x 2x

1. 52 = 4 • 13 = 4 • 13 = 2 13

2. 200 = 100 • 2 = 100 • 2 = 10 2

3. 4 54 = 4 9 • 6 = 4 • 9 • 6 = 4 • 3 • 6 = 12 6

4. 125x2 = 25 • 5 • x2 = 25 • 5 • x2 = 5x 5

Simplify 4 3 + 3.


= (4 + 1) 3 Use the Distributive Property to combine like radicals.

= 5 3 Simplify.

11-4

4 3 + 3 = 4 3 + 1 3 Both terms contain 3.

8 5 – 45 = 8 5 + 9 • 5 9 is a perfect square and a factor of 45.

Simplify 8 5 – 45.


= 8 5 – 9 • 5 Use the Multiplication Property of Square Roots.

= 8 5 – 3 5 Simplify 9.

= (8 – 3) 5 Use the Distributive Property tocombine like terms.

= 5 5 Simplify.

11-4


Simplify 5( 8 + 9).


5( 8 + 9) = 40 + 9 5 Use the Distributive Property.

= 4 • 10 + 9 5 Use the Multiplication Property of Square Roots.

= 2 10 + 9 5 Simplify.

11-4

Simplify ( 6 – 3 21)( 6 + 21).


( 6 – 3 21)( 6 + 21)

= 36 + 126 – 3 126 – 3 441 Use

FOIL.= 6 – 2 126 – 3(21) Combine like radicals and

simplify 36 and 441.

= 6 – 2 9 • 14 – 63 9 is a perfect square factor of 126.

= 6 – 2 9 • 14 – 63 Use the Multiplication Property of Square Roots.

= 6 – 6 14 – 63 Simplify 9.

= –57 – 6 14 Simplify.

11-4


= 2( 7 + 3) Divide 8 and 4 by the common factor 4.

= 2 7 + 2 3 Simplify the expression.

= Multiply in the denominator. 8( 7 + 3)

7 – 3

= Simplify the denominator. 8( 7 + 3)

4

11-4

Simplify . 8

7 – 3

= • Multiply the numerator and denominator by the conjugate of the denominator.

8

7 – 3

7 + 3

7 + 3

A painting has a length : width ratio approximately equal to the

golden ratio (1 + 5 ) : 2. The length of the painting is 51 in. Find the

exact width of the painting in simplest radical form. Then find the

approximate width to the nearest inch.


11-4

Define: 51 = length of painting x = width of painting

Relate: (1 + 5) : 2 = length : width

Write: =

x (1 + 5) = 102 Cross multiply.

= Solve for x. 102

(1 + 5)

x(1 + 5)

(1 + 5)

51 x

(1 + 5) 2


(continued)

x = Multiply in the denominator.102(1 – 5) 1 – 5

x = Simplify the denominator.102(1 – 5) –4

x = Divide 102 and –4 by the common factor –2.

– 51(1 – 5) 2

x = 31.51973343 Use a calculator.

x 32The exact width of the painting is inches.

The approximate width of the painting is 32 inches.

– 51(1 – 5) 2

11-4

x = • Multiply the numerator and the denominator by the conjugate of the denominator.

(1 – 5)

(1 – 5) 102

(1 + 5)




1. 5 6

2. 18 10

3. –2 5

4. 2 7

5. 14 2

6. –8 3

7. yes

8. yes

9. no

10. 4 2

11. –3 3

12. 4 2

13. –2 5

14. 7

15. 8 10

16. 4 – 4 2

17. 9 + 3

18. 6 – 2 3

19. 3 5 + 2 3

20. 3 2 + 6

21. 6 – 5 6

22. –9 – 14 6

23. 58 – 10 30

24. 11 – 4 7

25. 43 + 4 30

26. 32 + 9 11

27. 23 – 5 13

28. 2 7 + 2 3

29. –6 2

30. –4 6 – 12 2

31.

32. –5 11 – 5 3

33. 18 3 + 9 11

34. 10 2 + 10; 24.1

11-4

3( 10 + 5)5

13 + 65 + 130 + 5 28



35. – ; –1.3

36. 6 – 4 2; 0.3

37. 7.4 ft

38. 5 10

39. 6 2 + 6 3

40. 22 3 – 6

41. 8 + 2 15

42.

43. 15 + 4 14

44. –24

45. – 2

43

11-4

10 5

46. 4 3 + 4 2 + 3 6 + 6

47.

48. 8 2 units

49. 10 + 10 2 units

50. 6 10 units

51. 4x + x 10 units

52. Answers may vary. Sample: 8 2 + 4 3, 2 7 + 9 3, 6 5 + 3 7

53. a. The student simplified 48as 2 24 instead of 2 12 or 4 3.

b. 2 6 + 4 354. a. 2 2 or 2.8 ft

b. s 2

55. 9.1%

56. 12.8%

57. 15.5%

58. a. xb. x x

59.

60. about 251 years

61. They are unlike radicals.62. a. 1, 0, 1, 1;

4, 1, 5, 17; 5, 3, 8, 34; 8, 6, 14, 10; 10, 9, 19, 181

b. No; the only values it worked for were 0 and 1.



63. a + b = a + b

64.

65.

66.

67. 2

68. 10 2

69. 70

70. 2 2 – 6 – 3 + 3

71. a. 2 6b. 2 13c. 2(p + q)

72. B

n

2n – 1

2 ab b

9 22

8 1515

23 721

11-4

/



73. I

74. [2] (3 5 – 2)( 5 + 5 2) = 3 25 + 15 10 – 10 – 5 4 = 3(5) + 15 10 – 10 – 5(2) = 15 + 15 10 – 10 – 10 = 5 + 14 10

[1] correct technique, but with a computational error

75. [4]

[3] correct steps but answer not completely simplified

[2] correct technique, but with a computational error

[1] correct answer but no work shown

• Multiply the numerator and denominator by the conjugate of the denominator. Simplify the denominator.

11-4

7 – 21

7 – 21

5

7 + 21



76. 9.2 units

77. 6.7 units

78. 26.2 units

79. (3, 5)

80. (–2, 6.5)

81. 0, 7

82. –2, 9

83. –9, –3

84. –4, 6

85. –15, –2

86. –3, –

87. b2 + 22b + 121

88. 4p2 + 28p + 49

89. 25g2 – 49

90. 9x2 – 1

91. k2 – 81

92. d 2 – 2.2d + 1.21

12

19

11-4

16

5 – 7


1. 12 16 – 2 16 2. 20 – 4 5 3. 2( 2 + 3 3)

4. ( 3 – 2 21)( 3 + 3 21) 5.


40 –2 5 2 + 3 6

–123 + 3 7 –8 5 – 8 7

11-4

Solving Radical EquationsSolving Radical Equations



Evaluate each expression for the given value.

1. x – 3 for x = 16 2. x + 7 for x = 9 3. 2 x + 3 for x = 1


4. ( 3)2 5. ( x + 1) 2 6. ( 2x – 5) 2

11-5

Solving Radical EquationsSolving Radical EquationsALGEBRA 1 LESSON 11-5ALGEBRA 1 LESSON 11-5

Solutions

11-5

1. x – 3 for x = 16: 16 – 3 = 4 – 3 = 1

2. x + 7 for x = 9: 9 + 7 = 16 = 4

3. 2 x + 3 for x = 1: 2 1 + 3 = 2 4 = 2 • 2 = 4

4. ( 3)2 = 3

5. ( x + 1)2 = x + 1

6. ( 2x – 5)2 = 2x – 5

Solve each equation. Check your answers.


a. x – 5 = 4

x = 9 Isolate the radical on the left side of the equation.

( x)2 = 92 Square each side.

x = 81

Check: x – 5 = 4 – 5 4 Substitute 81 for x. 9 – 5 4 4 = 4

11-5

b. x – 5 = 4


x – 5 = 9 Solve for x.

x = 21

( x – 5)2 = 42 Square each side.

(continued)

Check: x – 5 = 4 21– 5 = 4 Substitute 21 for x. 16 = 4

4 = 4

11-5

On a roller coaster ride, your speed in a loop depends on the

height of the hill you have just come down and the radius of the loop in

feet. The equation v = 8 h – 2r gives the velocity v in feet per second

of a car at the top of the loop.


11-5


The loop on a roller coaster ride has a radius of 18 ft. Your car has a velocity of 120 ft/s at the top of the loop. How high is the hill of the loop you have just come down before going into the loop?

Solve v = 8 h – 2r for h when v = 120 and r = 18.120 = 8 h – 2(18) Substitute 120 for v and 18 for r.

= Divide each side by 8 to isolate the radical.

15 = h – 36 Simplify.

8 h – 2(18) 8

120 8

(15)2 = ( h – 36)2 Square both sides.225 = h – 36261 = h

The hill is 261 ft high.

(continued)

11-5

Solve 3x – 4 = 2x + 3.


( 3x – 4)2 = ( 2x + 3)2 Square both sides.

3x – 4 = 2x + 3 Simplify.

3x = 2x + 7 Add 4 to each side.

x = 7 Subtract 2x from each side.

The solution is 7.

Check: 3x – 4 = 2x + 3

3(7) – 4 2(7) + 3 Substitute 7 for x.

17 = 17

11-5


(x)2 = ( x + 12)2 Square both sides.

x2 = x + 12

x2 – x – 12 = 0 Simplify.

The solution to the original equation is 4. The value –3 is an extraneous solution.

Solve x = x + 12.

(x – 4)(x + 3) = 0 Solve the quadratic equation by factoring.

(x – 4) = 0 or (x + 3) = 0 Use the Zero–Product Property. x = 4 or x = –3 Solve for x.

Check: x = x + 12

4 4 + 12 –3 –3 + 12

4 = 4 –3 = 3 /

11-5


Solve 3x + 8 = 2.

3x = –6

( 3x)2 = (–6)2 Square both sides.

3x = 36

x = 12

3x + 8 = 2 has no solution.

Check: 3x + 8 = 2

3(12) + 8 2 Substitute 12 for x.

36 + 8 2

6 + 8 = 2 x = 12 does not solve the original equation./

11-5




1. 4

2. 49

3. 36

4. 137

5. 15

6. 16

7. 576 ft

8. 602 watts

9. 4.5

10. 3

11. 7

12. –2

13. 4

14. 2.5

15. 2

16. –4

17. none

18. –

19. –7

20. none

21. 3

22. 5

23. no solution

12

24. 2

25. no solution

26. 4

27. 1.25 or

28. , 1

29. a. 25b. 11.25

30. about 2.5 in.31. An extraneous solution is

a solution of a new equation that does not satisfy the original equation.

32. Answers may vary. Sample:

x – 2 = 7 – 2x , 3x = 3

14

11-5

54



33. 1600 ft

34. 3

35. no solution

36. no solution

37. 1, 6

38. 1.5

39. 11

40. 0, 12

41. 3, 6

42. 44

43. no solution

44. a. 68 ftb. 20.5 mi/hc. As radius increases,

velocity decreases. As height decreases, velocity decreases.

d. Velocity depends upon the difference of the height and the radius.

45. a.

b. approximately (6, 3.6)c. 6; it is the x-coordinate of

the point of intersection.

46. a. V = 10x2

b. x =

c. 2, 3, 4, 5, 6, 7

47. a. – 7, 7 b. 49c. In both cases 3 is

added to each side. To solve the first equation you find the square roots of each side, and in the second equation you find the square of each side.

48. –2, 8

49. 0

V10

11-5



51. –1

52. Subtract 2x from each side. Square both sides. Solve for x. Check the solution if there is one.

53. The square of x – 1 will have only 2 terms while x – 1 squared will have 3 terms.

54. a. about 2.0 mb. about 32.4 m

55. C

56. G

57. B

58. A

59. B

60. C

61. [2] 15 – 5x = 4x – 315 – 5x = 4x – 3

–9x = –18x = 2

Check: 15 – 5(2) 4(2) – 3

5 = 5The solution is 2.

[1] correct technique with a minor error OR correct answer, no work shown

62. 5 5

63. 3 2 + 4 3

64. 2

65. 32

11-5



66. 4( 5 – 3)

67. 54 2

68. 2.5, –4.5

69. 8.4, –0.4

70. –0.2, –4.8

71. –10.7, 0.7

72. –11.7, 1.7

73. –1.6, 3.1

74. (x + 12)(x – 2)

75. (m – 13)(m – 1)

76. (b + 18)(b – 2)

77. (2p + 1)(p + 7)

78. 3(d – 1)(d + 5)

79. (4v – 5)(v – 5)

11-5


Solve each radical equation.

1. 7x – 3 = 4 2. 3x – 2 = x + 2

3. 2x + 7 = 5x – 8 4. x = 2x + 8

5. 3x + 4 + 5 = 3

2

5 4

no solution

2 57

11-5

Graphing Square Root FunctionsGraphing Square Root Functions



Graph each pair of quadratic functions on the same graph.

1. y = x2, y = x2 + 3 2. y = x2, y = x2 – 4

Evaluate each expression for the given value of x.

3. x for x = 4

4. x + 7 – 3 for x = 2

5. 3 x + 2 for x = 9

11-6

Graphing Square Root FunctionsGraphing Square Root FunctionsALGEBRA 1 LESSON 11-6ALGEBRA 1 LESSON 11-6

Solutions

1. y = x2, y = x2 + 3

2. y = x2, y = x2 – 4

11-6

3. x for x = 4: 4 = 24. x – 7 – 3 for x = 2: 2 + 7 – 3 = – 3 = 9 – 3 – 3 = 05. 3 x + 2 for x = 9: 3 9 + 2 = 3(3) + 2 = 9 + 2 = 11

Find the domain of each function.


a. y = x + 5

The domain is the set of all numbers greater than or equal to –5.

b. y = 6 4x – 12

The domain is the set of all numbers greater than or equal to 3.

11-6

x –5>–

4x 12>–

x 3>–

x + 5 0 Make the radicand 0.>– >–

4x – 12 0 Make the radicand 0.>– >–

The size of a television screen is the length of the screen’s

diagonal d in inches.The equation d = 2A estimates the length of a

diagonal of a television with screen area A.


0 050

100200300400

1014.12024.528.3

ScreenArea(sq. in.)

Length ofDiagonal(in.)

11-6

Graph the function.

Domain2A 0

A 0

>–

>–


Graph y = x + 4 by translating the graph ofy = x .

For the graph y = x + 4,

the graph of y = x is shifted 4 units up.

11-6

For the graph ƒ(x) = x + 3,

the graph of y = x is shifted to the left 3 units.


Graph ƒ(x) = x + 3 by translating the graph ofy = x .

11-6

10. x y0 02 2

4.5 3

11. x ƒ(x) 0 0 1 2 4 4

12. x y2 03 26 4


1. x 2

2. x

3. x 0

4. x –7

5. x –3

6. x 5

7. x –

8. x –2

9. x



13. x y0 0 3 3

5.3 4

14. x ƒ(x)0 01 34 6

15. x y0 01 –34 –6

34

53

43

11-6

>–

>–

>–

>–

>–

>–

>–

>–

>–



16. h v0 01 84 16

17. D

18. A

19. C

20. B

21.

22.

23.

24.

25.

26.

27.

11-6



28.

29.

30. x 4; y 0

31. x 4; y 0

32. Form an inequality setting the radicand 0. Solve for x. Answers may vary. Sample:

y = x – 2 Domain: x – 2 0

x 2

11-6

33. a-d. Answers may vary. Samples:a. y = x + 2b. y = x + 2c. y = 2 xd. Check students’ work.

34. Translate the graph of y = x 8 units to the left.

35. Translate the graph of y = x 10 units down.

36. Translate the graph of y = x 12 units up.

37. Translate the graph of y = x 9 units right.

>–

>– >–

<–

>–>–

>–



38. x y2.5 03.5 16.5 2

39. x ƒ(x)0 01 4 2 5.7 4 8

40. x y–6 0–5 1–2 20 2.4

41. x y0 02 14 1.48 2

42. x y2 33 46 5

43. x ƒ(x)–2 –4–1 –32 –2

11-6



44. x y0 31 4.42 53 5.4

45. x y –3 1 –2 2.4 –1 3 0 3.4

46. x y 1 –2 2 –0.3 3 0.4 4 1

47. B

48. D

49. A

50. C

51. a. p > 0b.

c. about 45 lb/in.2

52. a. nob. Answers may vary.

Sample: The graph of y = x is the first quadrant of the graph of x = y2.

c. y = – x

53. y = 3 x rises more steeply because 3 x > 3x for all positive values of x.

54. False; x must equal 81.

55. False; only combine like terms.

56. true

57. False; x = –1.

11-6



58. a. about 213 camerasb. month 4

59. a.

b. y = |x| + 5

60. Translate the graph of y = x right 2 units and up 3 units.

61. a. i. ii.

iii. iv.

b. The greater the absolute value of n, the steeper the graph. If n < 0, then the graph lies in Quadrant II. If n > 0, the graph lies in Quadrant I.

11-6



62. Check students’ work.

63. B

64. H

65. C

66. H

67. B

68. H

69. [2] x y6 07 18 1.49 1.7

10 2[1] incorrect coordinates on graph

70. 16

71. 7

72. 169

73. 14.76

74. no solution

75.

76. ,

77. 4 – 39, 4 + 39

78. no solution

79. ,

23

–2 – 3 22

–2 + 3 22

–1 – 113

–1 + 113

11-6

80. no solution

81. ,

82. (2x + 1)(x – 4)

83. (3x – 5)(x + 2)

84. (2x + 1)(2x + 9)

85. 2(x – 8)(x + 3)

86. 4(x2 – x – 15)

87. x(x – 13)(x + 1)



–13 – 42118

–13 + 42118

11-6

x > 2


Shift the graph to the right 15 units.

11-6

1. Find the domain of the function ƒ(x) = 2x – 4.

2. Graph y = 3 x.

3. Graph y = x – 3.

4. Describe how to translate the graph of y = x

to obtain the graph of the function y = x – 15.

Trigonometric RatiosTrigonometric Ratios



11-7

Let c = , s = , t = . Calculate c, s, and t for the given values.

1. A = 3, O = 4, H = 5 2. A = 5, O = 12, H = 13

Solve each equation.

3. = 4. =

5. = 6. =

AH

OH

OA

15 x

0.75 1

0.84 1

21 x

x 20

x 0.52

0.34 1

14 1

Trigonometric RatiosTrigonometric RatiosALGEBRA 1 LESSON 11-7ALGEBRA 1 LESSON 11-7

Solutions

1. For A = 3, O = 4, H = 5:

c = = , s = = , t = =

2. For A = 5, O = 12, H = 13:

c = = , s = = , t = =

3. = 4. =

0.75x= 15 x = 20(0.34)

x = 20 x = 6.8

5. = 6. =

0.84 = 21 x = 14(0.52)

x = 25 x = 7.28

AH

OH

OA

AH

OH

OA

35

45

34

5 13

1213

12 5

15 x

0.75 1

x 20

0.34 1

0.84 1

21 x

x 0.52

14 1

11-7

1. For A = 3, O = 4, H = 5:

c = = , s = = , t = =

2. For A = 5, O = 12, H = 13:

c = = , s = = , t = =

3. = 4. =

0.75x= 15 x = 20(0.34)

x = 20 x = 6.8

5. = 6. =

0.84 = 21 x = 14(0.52)

x = 25 x = 7.28

AH

OH

OA

AH

OH

OA

35

45

34

5 13

1213

12 5

15 x

0.75 1

x 20

0.34 1

0.84 1

21 x

x 0.52

14 1

Use the triangle. Find sin A, cos A, and tan A.


sin A = = =opposite leghypotenuse

6 10

35

tan A = = =opposite legadjacent leg

68

34

cos A = = = 8 10

45

adjacent leghypotenuse

11-7


Find sin 40° by using a calculator.


Rounded to the nearest ten-thousandth, the sin 40°is 0.6428.

Use degree mode when finding trigonometric ratios.

To find sin 40°, press 40 .

11-7

Find the value of x in the triangle.


Step 1: Decide which trigonometric ratio to use.

You know the angle and the length of the hypotenuse.You are trying to find the adjacent side. Use the cosine.

Step 2: Write an equation and solve.

x = 15(cos 30°) Solve for x.

cos 30° = adjacent leghypotenuse

cos 30° = Substitute x for adjacent leg and 15 for hypotenuse.

x 15

x 13.0 Round to the nearest tenth.15 30 12.99038106 Use a calculator.

The value of x is about 13.0.

11-7

Suppose the angle of elevation from a rowboat to the top of a

lighthouse is 708. You know that the lighthouse is 70 ft tall. How far

from the lighthouse is the rowboat? Round your answer to the nearest

foot.


Define: Let x = the distance from the boat to the lighthouse.

Relate: You know the angle of elevation and the opposite side. You are trying to find the adjacent side. Use the tangent.

Draw a diagram.

11-7

x(tan 70°) = 70 Multiply each side by x.

x = Divide each side by tan 70°. 70tan 70


(continued)

Write: tan A =

tan 70° = Substitute for the angle and the sides.

opposite legadjacent leg70 X

The rowboat is about 25 feet from the lighthouse.

x 25.4779164 Use a calculator.

x 25 Round to the nearest unit.

11-7

A pilot is flying a plane 15,000 ft above the ground. The pilot

begins a 3° descent to an airport runway. How far is the airplane from

the start of the runway (in ground distance)?


Draw a diagram.

Define: Let x = the ground distance from the start of the runway.

Relate: You know the angle of depression and the opposite side. You are trying to find the adjacent side. Use the tangent.

11-7


(continued)

x(tan 3°) = 15,000 Multiply each side by x.

x = Divide each side by tan 3°.15,000 tan 3

Write: tan A =

tan 3° = Substitute for the angle and the sides.15,000 x

opposite legadjacent leg

x 286217.05 Use a calculator.

x 290,000 Round to the nearest 10,000 feet.

The airplane is about 290,000 feet (or about 55 miles) from the start of the runway.

11-7




1.

2.

3.

4.

5.

6.

7. 0.5299

8. 0.5736

9. 1.2799

10. 0.9962

11. 0.9659

354534453543

12. 5.5

13. 10.4

14. 19.2

15. 38.1

16. 66.0

17. 21.1

18. about 2.0 mi

19. about 172 ft

20. about 816 ft

21. about 0.4 mi

22. c = 29; sin A = ;

cos A = ; tan A =

2129

2029

2120

23. b = 15; sin A = ;

cos A = ; tan A =

24. a = 24; sin A = ;

cos A = ; tan A =

25. AC 6; AB 8

26. AC 36; BC 22

27. BC 6; AB 18

28. AC 4; AB 50

29. about 55 m

30. about 4.4 m

31. a. 1,720,000 ftb. 326 mi

8 17

1517

815

1213

5 13

125

11-7

40. Answers may vary. Sample: about 8.2 cm

41. about 203 ft

42. about 5938 m

43. about 57 ft

44. 550 ft

45. 0.25

46. a. about 229 kmb. about 4 km

47. C

48. G



32. 12 m

33. about 6.8 m

34. 514.3

35. 4.5

36. 78.4

37. q 6.1; r 7.9

38. a. about 177 ft

b. about 86 ft

39. a. about 252 ft

b. about 377 ft

49. [2] sin A 0.6271, cos A 0.7881, tan A 0.7957

[1] at least one correct equation

50. [4]

tan 56° =

x 573.3573 ft

[3] correct equation, but minor computational error

[2] incorrect equation used[1] no work shown

850x

11-7



51.

52.

53.

54. 2

55. 0

56. 1

57. (n – 20)(n + 20)

58. (x – 15)2

59. (10p – 7)(10p + 7)

60. d – d +

61. 2(7w – 8)(7w + 8)

62. (x + 13)2

14

32

14

32

11-7

14

1. Use the figure to find sin A, cos A, and tan A.

2. Find the value of x to the nearest tenth.

3. A group of skateboarders wants to build a ramp with anangle of incline of 14°. What should the rise be for every10 meters of run?

4. A park ranger on a 220 ft tower spots a fire at an angleof depression of 4°. How far is the fire from the base ofthe tower?

5. A wheelchair ramp is to have an angle of 3.5° with the ground. The deck at the top of the ramp is 18 in. above the ground. How long should the ramp be?


2129

2029

2120

, ,

6.9

about 2.5 m

about 3146 ft

about 294.3 in.

11-7

12. – , – 2

13. – , 5 (– , 5)

14. 29.5

15. 27.0

16. 13 ft

17. 15 mi

18. 2

19.

20. 4 3

21. 4 6

22. 17 2

ALGEBRA 1 CHAPTER 11ALGEBRA 1 CHAPTER 11

Radical Expressions and EquationsRadical Expressions and Equations

1. yes

2. no

3. no

4. yes

5. 57.8 cm

6. 9.8 units

7. 4.1 units

8. 19.8 units

9. 7.6 units

10. 2 , 2

11. , – 3

12

12

12

12

12

12

35

23. 11 3

24. 35 5 – 15 7

25. 8 2 – 8 3

26. 5 3

27. 2 10 – 6

28. Answers may vary. Sample: 2 5 + 4 5 = 6 5

29. B

30. 4

31. 13

32. 6

33. 7

11-A

ALGEBRA 1 CHAPTER 11ALGEBRA 1 CHAPTER 11

Radical Expressions and EquationsRadical Expressions and Equations

11-A

34.

35. 10.65

36. 2 5 ft

37. x 0;

95

38. x 0;

39. x 4;

40. x –9;

41. 24 cm

42. The graph of y = x is shifted 3 units down.

43. r =

44. about 32 ft

45. about 8.1

46. about 12.1

47. about 1.106

48. about 0.7431

49. BC 6.2, AC 3.3

50. about 5 ft

V h

>–

>–

>–

>–

simplifying radicals (for help, go to lessons 8-3 and 10-3.) algebra 1 lesson 11-1 complete each...

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