Roots and Radical Expressions

Roots and Radical ExpressionsRoots and Radical ExpressionsALGEBRA 2 LESSON 7-1ALGEBRA 2 LESSON 7-1

(For help, go to Lesson 5-4.)

Write each number as a square of a number.

1. 25 2. 0.09 3.

Write each expression as a square of an expression.

4. x10 5. x4y2 6. 169x6y12

449

7-1


1. 25 = 52 2. 0.09 = 0.32 3. = = 2

4. x10 = (x5)2 5. x4y2 = (x2y)2 6. 169x6y12 = (13x3y6)2

449

22

72

27

Solutions

7-1


Find all the real roots.

Since 0.33 = 0.027, 0.3 is the cube root of 0.027.

Since (–5)3 = –125, –5 is the cube root of –125.

Since 54 = 625 and (–5)4 = 625, 5 and –5 are fourth roots of 625.

There is no real number with a fourth power of –0.0016.

Since 3 = , is the cube root of .

14

164

14

164

b. the fourth roots of 625, –0.0016, and 81

625

Since 4 = and

4 = , and – are fourth roots of .

81625

35

81625

–35

81625

35

35

7-1

a. the cube root of 0.027, –125, and 1 64


Find each real-number root.

3a. –1000

There is no real number whose square is –81.

Rewrite –1000 as the third power of a number.3= (–10)3

Definition of nth root when n = 3, there is only one real cube root.

= –10

b. –81

7-1

3b. a3b3


Simplify each radical expression.

Absolute value symbols must not be used here. If a or b is negative, then the radicand is negative and the root must also be negative.

a. 9x10

Absolute value symbols ensure that the root is positive when x is negative.

9x10 = 32(x5)2 = (3x5)2 = 3| x5|

(ab)3 = ab3

7-1


(continued)

Absolute value symbols ensure that root is positive when y is negative. They are not needed for x because x2 is never negative.

c. x16y44

4 4x16y14 = (x4)4(y)4 = x4 |y|

7-1


A cheese manufacturer wants to ship cheese balls that weigh

from 10 to 11 ounces in cartons that will have 3 layers of 3 cheese

balls by 4 cheese balls. The weight of a cheese ball is related to its

diameter by the formula w = , where d is the diameter in inches and

w is the weight in ounces. What size cartons should be used? Assume

whole-inch dimensions.

d3

5

Find the diameter of the cheese balls.

7-1

<–10 w 11 Write an inequality.

10 11 Substitute for w in terms of d.d3

5

<–

<– <–

50 d3 55 Multiply by 5.<– <–

3 3 3 50 d3 55 Take cube roots.

3.68 d 3.80 The diameters range from 3.68 in. to 3.80 in.

<– <–

<– <–


(continued)

The length of a row of 4 of the largest cheese balls is 4(3.80 in.) = 15.2 in.

The length of a row of 3 of the largest cheese balls 3(3.80 in.) = 11.4 in.

The manufacturer should order cartons that are 16 in. long by 12 in. wide by 12 in. high to accommodate three dozen of the largest cheese balls.

7-1


pages 366–367 Exercises

1. 15, –15

2. 0.07, –0.07

3. none

4. , –

5. –4

6. 0.5

7. –

8. 0.07

9. 2, –2

10. none

11. 0.3, –0.3

8 13

8 13

12

12. , –

13. 6

14. –6

15. no real-number root

16. 0.6

17. –4

18. –4

19. –3

20. no real-number root

21. 4|x|

22. 0.5|x3|

23. x4|y9|

103

103

24. 8b24

25. –4a

26. 3y2

27. x2|y3|

28. 2y2

29. 1.34 in.

30. 1.68 ft

31. 0.48 cm

32. 0.08 mm

33. 10, –10

34. 1, –1

35. 0.5, –0.5

7-1


36. , –

37. –64, 64, – –64, 64

38. a. 35 ftb. 20 ft longer

39. 0.5

40.

41. 0.2

42.

43. 2|c|

44. 3xy2 3

45. 12y2z2x xy

46. y4

23

23

13

14

3 36

3

47. –y4

48. k3

49. –k3

50. |x + 3|

51. (x + 1)2

52. |x|

53. x2

54. |x3|

55. Answers may vary. Sample:

–8x6, – 16x8, –32x10

56. a. for all positive integersb. for all odd positive integers

3 4 5

7-1


57. Yes, because 10 is really 101.

58. All; x2 is always nonnegative.

59. Some; true only for x > 0.

60. Some; true only for x = –1, 0, 1.

61. Some; true only for x > 0.

62. |m|

63. m2

64. |m3|

65. m4

66. m

67. m2

68. m3

69. m4

70. B

71. I

72. B

73. H

74. [2] x2y4 equals x3y6 whenever y = 0, regardless of the value of x. This is true because x2y4 and x3y6 will be 0 whenever y = 0. x2y4 also equals x3y6 when x > 0. This is true because x2y4 = |x|y2 and x3y6 = xy2, and |x|y2 = xy2 when x > 0.

[1] answer only, with no explanation

3

3

3

7-1


75. x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5

76. 16 – 96y + 216y2 – 216y3 + 81y4

77. 729x6 – 7290x5 + 30,375x4 – 67,500x3 + 84,375x2 – 56,250x + 15,625

78. 128a7 – 448a6b + 672a5b2 – 560a4b3 + 280a3b4 – 84a2b5 + 14ab6 – b7

79. y = x(2x – 7)(2x + 7)

80. y = (9x + 2)2

81. y = 4x(x + 1)2

82. y = 2x(6x + 1)(x + 1)

83. y = 3(x – 0)2 – 7

84. y = –2 – x – –

85. y = (x + 4)2 – 5

14

2 798

7-1

14

13

3


125

1216

± 11 none ± 8 none

–20 16

none0.7 5 –9

–2x 4y2 6 | x7 |

about 3.37 cm

7-1

3 4

1. Find all the real square roots of each number.

a. 121 b. –49 c. 64 d. –

2. Find all the real cube roots of each number.

a. –8000 b.

3. Find each real-number root.

a. 0.49 b. 125 c. – 81 d. –6254. Simplify each radical expression.

a. –8x3 b. 16y4 c. 36x14

5. The formula for the volume of a cone with a base of radius r and height

r is V = r 3. Find the radius to the nearest hundredth of a centimeter

if the volume is 40 cm3.

Multiplying and Dividing Radical ExpressionsMultiplying and Dividing Radical ExpressionsALGEBRA 2 LESSON 7-2ALGEBRA 2 LESSON 7-2

(For help, go to page 362.)

Find each missing factor.

1. 150 = 52( ) 2. 54 = ( )3(2)

3. 48 = 42( ) 4. x5 = ( )2(x)

5. 3a3b4 = ( )3(3b) 6. 75a7b8 = ( )2(3a)

7-2


1. 150 = 52(6) 2. 54 = (3)32

3. 48 = 42(3) 4. x5 = (x2)2(x)

5. 3a3b4 = (ab)3(3b) 6. 75a7b8 = (5a3b4)2(3a)

Solutions

7-2


Multiply. Simplify if possible.

b. –16 • 4

a. 3 • 12

3 • 12 = 3 • 12 = 36 = 6

The property for multiplying radicals does not apply. –4 is not a real number.

c. –4 • 16

3 3

7-2

3 333–16 • 4 = –16 • 4 = 64 = –4

a. 50x5


Simplify each expression. Assume all variables are positive.

= 5x2 2x definition of nth root

Factor into perfect squares.50x5 = 52 • 2 • (x2)2 • x

3b. 54n8

33 54n8 = 33 • 2 • (n2)3 • n2 Factor into perfect cubes.

3= 3n2 2n2 definition of nth root

7-2

33= 33(n2)3 • 2n2 n n na • b = ab

= 52 • (x2)2 • 2 • x n n na • b = ab

33 25xy8 • 5x4y3


Multiply and simplify . Assume all variables

are positive.

3= 53x3(y3)3 • x2y2 Factor into perfect cubes.

3= 5xy3 x2y2 definition of nth root

7-2

3 3= 53x3(y3)3 • x2y2 n n na • b = ab

25xy8 • 5x4y3 = 25xy8 • 5x4y33 3 3 n n na • b = ab

3= 64x7


Divide and simplify. Assume all variables are positive.

a.

3= –27

3= (–3)3

= –3

3 –81

33=

b.

3x3

3 192x8

=

192x8

3x

3

=

= 4x2 x3

3= 43(x2)3 • x

7-2

=3 –81

3


Rationalize the denominator of each expression. Assume that

all variables are positive.

155=

a.

5

3

7-2

Method 1:

Rewrite as a square root of a fraction.

35

=5

3

= 3 • 55 • 5

Then make the denominator a perfect square.

= 1552


(continued)

3 • 55 • 5

Multiply the numerator and denominator by 5 so the denominator becomes a whole number.

=3

5

Method 2:

a.

155=

7-2

5

3


(continued)

b. x5

3x2y

c. 3 54y

x5

3x2y

x5 3x2y

3x2y3x2y

•

•=

3x7y3x2y

=

x3 3xy3x2y

=

x 3xy3y

=

3 5 • 42y2

4y • 42y2

3 54y

=

Rewrite the fraction so the denominator is a perfect cube.

3 80y2

43y3=

2 10y2

4y3

=

10y2

2y

3=

7-2


The distance d in meters that an object will fall in t seconds is

given by d = 4.9t 2. Express t in terms of d and rationalize the

denominator.

d = 4.9t 2

t 2 =d

4.9

t =d

4.9

10d7

=

d • 1049=

7-2



1. 16

2. 4

3. –9

4. 4

5. not possible

6. not possible

7. –6

8. 6

9. 2x 5x

10. 3 3x2

11. 5x2 2x

3

12. 2a 4a2

13. 3y3 2y

14. 10a3b3 2b

15. –5x2y 2y2

16. 2y 4x3y2

17. 2 12

18. 8y3 5y

19. 7x3y4 6y

20. 40xy 3

21. 30y2 2y

22. –2x2y 30x

23. 10

24.

25. 2x2y2 2

26. 5x3 x2y2

27.

28.

29.

30.

31.

32. 5x2 5

33.

4xy

3

3

3

3

4

3

3

2x2

10x4x

4x2

2505

15y5y

45x2

3x4

3

3

7-2


48.

49.

50.

51. –

52. –4 4 – 6 2

53.

54.

55. 212 mi/h greater

56. 20 22 cm2

7-2

34.

35. r =

36. a.

b.

c. Answers may vary. Sample: First simplify the denominator. Since 98 = 2 • 49 = 7 2, to rationalize the denominator,

multiply the fraction by .

This yields =

.

x 102y

Gm1m2F

F 6 + 3

15 6 + 3

15

2

2 2 • 2 + 3 • 2

7 2 • 22 + 6

14

37. 10 2

38. 4 5

39. 3x6y5 2y

40. 20x2y3 y

41. 10 + 7 2

42. 15 + 3 21

43. 5 + 5 3

44. 2x 2

45. 3x2 x

46.

47.

3

3

3

x 10y2y 2

5 14x21x

3 3

3x2

3x

3

6 – 24

3 5 + 55

33x4x

2xy 2

xy

3

2 25xx

3


x2yxy

3

7-2

57. A product of two square roots can be simplified in this way only if the square roots are real numbers. –2 and –8

are not.

58. 288a5 ft

59. For some values; it is easy to see that the equation is true if x = 0 or x = 1. But when x < 0, x3 is not a real number, although x2 is.

60. Check students’ work.

61. 2xy

62. 2xy2

63. 2 5

3

64.

65.

66.

67. a = –2c, b = –6d

68. No changes need to be made; since they are both odd roots, there is no need for absolute value symbols.

69. C

70. H

71. A

72. G

yx x

5

x4y3

y

6


7-2

3 2x

4x2

4x2

3

3

3 4x2

4x2

3

3

3

2x

3

3

12x2

8x3

3

3

12x2

2x

3

3 2x

3 4x2

4x2

3

3

<–

>–

73. D

74. [2] x is a real number if x 0 and –x is a real number if x 0. So the only value that makes x • –x a real number is x = 0.

[1] answer only OR error describing value(s) of variables

75. [4] You should multiply by because

• = • = = ,

which has a denominator without a radical.[3] appropriate methods, but with one minor error[2] major error, but subsequent steps consistent with that error[1] correct final expression, but no work shown


7-2

76. 11|a45|

77. –9c24d32

78. –4a27

79. 2y5

80. 0.5|x3|

81. x2y5

82. 2|x9|y24

83. 0.08x20

84. y2 – 4y + 16, R –128, not a factor

85. x2 – 3x + 9, a factor

86. 3a2 – a, R 4, not a factor

87. 2x3 + x2 + 2x, R 10, not a factor

88. 25

89. 25

90.

91.

92.

93. 0.0225

94.

95.

1214

1214

1 36

9 64 9 100

3

3 3

3 3

3

3 x2

4


128x3

2xy

7x3

3

3

270x

10xy2

15 20

2x2 2x 3–3xy3 9y

6x2y xy 32xy 5xy2

8x yy

33 yy

3 2x2

221x3

7-2

Assume that all variables are positive.

1. Multiply. Simplify if possible.

a. 5 • 45 b. 4 • 2000

2. Simplify.

a. 8x5 b. –243x3y10

3. Multiply and simplify.

a. 18x3 • 2x2y3 b. 10x2y4 • 4x2y

4. Divide and simplify.

a. b.

5. Rationalize the denominator of each expression.

a. b.

Binomial Radical ExpressionsBinomial Radical ExpressionsALGEBRA 2 LESSON 7-3ALGEBRA 2 LESSON 7-3

(For help, go to Lesson 5-1 or Skills Handbook page 853.)

Multiply.

1. (5x + 4)(3x – 2) 2. (–8x + 5)(3x – 7)

3. (x + 4)(x – 4) 4. (4x + 5)(4x – 5)

5. (x + 5)2 6. (2x – 9)2

7-3


1. (5x + 4)(3x – 2)= (5x)(3x) + (5x)(–2) + (4)(3x) + (4)(–2)= 15x2 – 10x + 12x – 8= 15x2 + 2x – 8

2. (–8x + 5)(3x – 7)= (–8x)(3x) + (–8x)(–7) + (5)(3x) + (5)(–7)= –24x2 + 56x + 15x – 35= –24x2 + 71x – 35

3. (x + 4)(x – 4) = x2 – 42 = x2 – 16

4. (4x + 5)(4x – 5) = (4x)2 – 52 = 16x2 – 25

5. (x + 5)2 = x2 + 2(5)x + 52 = x2 + 10x + 25

6. (2x – 9)2 = (2x)2 – 2(2x)(9) + 92 = 4x2 – 36x + 81

Solutions

7-3


Add or subtract if possible.

a. 7 xy + 3 xy

Distributive Property7 xy + 3 xy = (7 + 3) xy

= 10 xy Subtract.

3 3b. 2 x – 2 5

The radicals are not like radicals. They cannot be combined.

7-3


The rectangular window shown below is made up of three

equilateral triangles and two right triangles. The equilateral triangles

have sides of length 4 feet. What is the perimeter of the window?

So the window’s height is 2 3 ft.

The length of the window is 8 ft.

The perimeter is 2(2 3 + 8) ft or 16 + 4 3 ft.

The height of an equilateral triangle with side length 4 ft is 2 3 ft.

7-3

Simplify 3 20 – 45 + 4 80.


3 20 – 45 + 4 80 = 3 22 • 5 – 32 • 5 + 4 42 • 5 Factor each radicand.

Simplify each radical.= 3 • 2 5 – 3 5 + 4 • 4 5

Multiply.= 6 5 – 3 5 + 16 5

Use the Distributive Property.= (6 – 3 + 16) 5

= 19 5

7-3


Multiply (2 + 4 3)(1 – 5 3).

Use FOIL.(2 + 4 3)(1 – 5 3) = 2 • 1 – 2 • 5 3 + 4 3 • 1 – 4 3 • 5 3

= 2 + (4 – 10) 3 – 60 Distributive Property

= –58 – 6 3

7-3


Multiply (3 + 7)(3 – 7).

= 9 – 7 = 2

(a + b)(a – b) = a2 – b2(3 + 7)(3 – 7) = 32 – ( 7)2

7-3


Rationalize the denominator of2 – 3

4 + 3

2 – 3

4 + 3=

2 – 3

4 + 3•

4 – 3

4 – 34 – 3 is the conjugate of 4 + 3.

= •2 – 3

4 + 3

4 – 3

4 – 3Multiply.

42 – ( 3)2

8 – 2 3 – 4 3 + ( 3)2

= Simplify.

11 – 6 313

=

7-3



1. 6 6

2. 4 3

3. cannot combine

4. –2 x

5. cannot combine

6. 5 x2

7. 33 2

8. 13 5

9. 7 2

10. 5 2

11. 9 3 – 6 23 3

3

3

3

12. 2 2 + 2 3

13. 8 + 4 5

14. 23 + 7 7

15. 63 – 38 2

16. 8 + 2 15

17. 49 + 12 13

18. 38 + 12 10

19. 14

20. 4

21. –40

22. –2

23. –2 + 2 3

4 4 24.

25. 13 + 7 3

26.

27. 13 2

28. 8 3

29. 48 2x

30. 5 3 – 4 2

31. 33y 6

32. –2 2

33. –11 + 21

34. 8 + 10

35. 17 + 31 2

3

11 + 8 2–14

12 3 + 823

7-3


36. –36 – 15 2

37. x + 3 3x + 6

38. 8y – 22 2y + 30

39.

40. 2 3 – 2

41.

42.

43.

44. 2 2 – 12

45. The reciprocal is ,

which is one less than .

89 + 42 3–239

3 – 72

2 + 3 42

x + 5 x3

x

3

4

33

–1 + 52

1 + 52

46. a must be twice a perfect square.

47. 14 7 m

48. Answers may vary. Sample: Without simplifying first, you must estimate three separate square roots, and then add the estimates. If they are first simplified, then they can be combined as 13. Then only one square root need be estimated.

49. Answers may vary. Samples: ( 7 + 2)( 7 – 2); (2 2 + 5)(2 2 – 5)

50. s, or about 4.53 s

51. –

52. 4 3

60 – 20 27

12

7-3


53. (a = 0 and b > 0) or (b = 0 and a > 0)

54. In the second step the exponent was incorrectly distributed: (a – b)x ax – bx.

55. a. m and n can be any positive integers.b. m must be even or n must be odd.c. m must be even, and n can be any positive integer.

56. B

57. I

58. D

59. I

60. D

61. [2] (1 – 8)(1 + 8) = (1 – 2)(1 + 2) = 1 – 4 = –3, which is rational.

[1] the value –3 only OR yes only OR minor error in calculation

3 3

7-3

=/


62. [4] – =

• –

• =

– =

– =

=

[3] minor error, but appropriate method

[2] correct method, but finds

+ and gets

[1] answer only, no work shown

25 + 2 2

35 – 2 2

25 + 2 2

35 – 2 2

5 + 2 25 + 2 2

5 – 2 25 – 2 2

10 – 4 217

15 + 6 225 – 4 • 2

10 – 4 225 – 4 • 2

15 + 6 217

10 – 4 2 – 15 – 6 217

10 – 4 217

15 + 6 217

25 + 2 217

63. 3 2

64. x 15

65. 4

66.

67. 2x

68. 7x2 2

69.

70.

71. 2, –1 ± i 3

72. –10, 5 ± 5i 3

73. ,

–5 – 10 217

933

3nn

100x2

5x

15

–1 ± i 310

74. ± 7

75.

76. ± , ± i 3

13

±2 55

7-3

3

3


3 3

3 3

2 10 313 x

10 3 3 2

29 + 3 3

16 – 17 793

7-3

3 – 2 710 – 7

Simplify.

1. 6 10 – 4 10 2. 8 x + 5 x

3. 2 27 + 48 4. 128 – 54

5. Multiply (7 + 2 3) (5 – 3).

6. Rationalize the denominator in .

Rational ExponentsRational Exponents

(For help, go to page 362.)

Simplify.

1. 2–4 2. (3x)–2

3. (5x2y)–3 4. 2–2 + 4–1

5. (2a–2b3)4 6. (4a3b–1)–2

ALGEBRA 2 LESSON 7-4ALGEBRA 2 LESSON 7-4

7-4

Rational ExponentsRational ExponentsALGEBRA 2 LESSON 7-4ALGEBRA 2 LESSON 7-4

7-4

124

116

1(3x)2

19x2

1(5x2y)3

1125x6y3

122

141

14

14

24

12

16b12

a8

b2

42a6

b2

16a6

Solutions

1. 2–4 = =

2. (3x)–2 = =

3. (5x2y)–3 = =

4. 2–2 + 4–1 = + = + = =

5. (2a–2b3)4 = 24a–8b12 =

6. (4a3b–1)–2 = 4–2a–6b2 = =


Simplify each expression.

3= 43 Rewrite 64 as a cube.

= 4 Definition of cube root.

a. 641

3

364 = 64 Rewrite as a radical.1

3

= 7 By definition, 7 is the number whose square is 7.

b. 7 • 7 1

2

1

2

7 • 7 = 7 • 7 Rewrite as radicals.1

2

1

2

7-4


(continued)

c. 5 • 25 1

3

1

3

5 • 25 = 5 • 25 Rewrite as radicals.1

3

1

33 3

3= 5 • 25 property for multiplying radical expressions

7-4

= 5 By definition, 5 is the number whose cube is 5.

3


a. Write x and y –0.4 in radical form.2

7

y –0.4 = y = or 2

5

–1y25

1y 25

b. Write the radical expressions and in exponential form.c34 b 53

= c 3

4c34 b 53 = b

5

3

7-4

2

7x = x2 or x 2 7 7


The time t in hours needed to cook a pot roast that weighs p

pounds can be approximated by using the equation t = 0.89p0.6. To

the nearest hundredth of an hour, how long would it take to cook a

pot roast that weighs 13 pounds?

t = 0.89p0.6 Write the formula.

= 0.89(13)0.6 Substitute for p.

4.15 Use a calculator.

7-4


Simplify each number.

a. (–27) 2

3

Method 1

= (–3)2

= 9

(–27) = ((–3)3) 2

3

2

3

Method 2

3= ( (–3)3)2

= (–3)2 = 9

3(–27) = ( –27)2 2

3

7-4


(continued)

b. 25–2.5

Method 1

25–2.5 = 25 5

2

–

= (52) 5

2

–

= 525

2

–

= 5–5

=155

=1

3125

Method 2

25–2.5 = 25 5

2

–

1

255

2

=

=155

13125=

=1

( 25)5

7-4


Write (243a–10) in simplest form.

= 32a–4

2

5

(243a–10) = (35a–10) 2

5

2

5

= 35 • a(–10) 2

5

2

5

32

a4=

=9a4

7-4


7-4

7 7

5 5

55


1. 6

2. 3

3. 7

4. 10

5. –3

6. 6

7. 8

8. 3

9. 3

10. x

11. x5

6

1

2

1

2

3

2

3

2

3

2

1

y98

1

y8 9

1

t 34

1

t4 3

22. a

23. a

24. c

25. 25x2y2

26. 72.8 m

27. 15.1 m

28. 7.9 m

29. 1.6 m

30. 4

31. 16

32. 4

33. 64

1

2

2

32

3

12. x2 or ( x)2

13. y2 or ( y)2

14. or

15. or

16. x3 or ( x)3

17. y6 or ( y)6

18. (–10)

19. 7 x

20. (7x)

21. (7x)


7-4

34.

35. 8

36. 64

37. 1000

38.

39.

40.

41.

42. –

43. –2y3

1 16

1 x2

1 x4

1

3x 2

3 5

x 2

3 3 x3

44.

45. x

46.

47.

48. x3y9

49.

50. –7

51. –3

52. 64

53. 729

1x

y4

x3

y2

x8

y5

x10

13

3

54. 2,097,152

55. 1,000,000,000 or 109

56.

57.

58.

59. 16

60. –

61. 10

62. 78%, 61%, 37%

63. 635.87

64. 768

1418 1 36

1 81


7-4

65. x

66. y

67. x

68. y

69. x y

70.

71.

72.

73.

1

2

4

5

1

2

1

2

1

6

1

4

1

x y1

4

5

64x7

9y9

9y8

4x6

1

x 13

36

74.

75.

76.

77. The cube root of –64 is –4,

which equals –(64) . The square root of –64 is not a real number, but

–(64) = – 64 = –8.

78. The exponent applies

only to the 5, not to the 25.

1

x 7

24 1

x 1

3 1

(xy) 1

2

1

3

1

2

1

2

1

2

4 – 5 2

1

2

1

4

1

2

2

4

12

4

79. a. Answers may vary.

Sample: 4 – 5 ,

2(4 – 5 ), b. no

80. a. x • x • x • x =

x • x = x2, so x2 = x

b. x2 = (x2) = x = x = x

81. 49

82. 9

83. x2

84. 1


85. 3 2

86. 9

87. 33.13 mi/h

88. B

89. H

90. [2] x =

[1] minor error

91. B

92. A

93. C

94. A

95. 4 3

1 16

96. 21 2

97. 1 + 3 5

98. 15 – 4 14

99.

100.

101. 4x(x2 – 2x + 4)

102. (x + 2)2

103. (x – 9)2

104. (4a – 3b)(4a + 3b)

105. (5x – 4y)2

106. (3x + 8)2

4 + 6 5 + 2 10 + 15 2–41

10 – 8 2 7

7-4

3


1. Simplify each expression.

a. 100 b. (–64)

2. Write each expression in radical form.

a. x b. y c. k1.8 d. a

3. A container with curved sides is 10 ft tall. When water is in the container to a depth of h ft, the number of cubic feet of water can be approximated by using the formula V = 0.95h2.9. Find the amount of water in the container when the depth of the water is 7.5 ft. Round to the nearest hundredth.

4. Simplify each number.

a. (–64) b. 810.75

5. Write x y in simplest form.

1

2

1

3

1

7

4

5

1

6–

4

3

10 –4

x7 1a6

327.64 ft3

256 27

y

x8

3

7-4

2

3

1

4– –4

y45 or y 54

k95 or k 59

Solving Radical EquationsSolving Radical Equations



Solve by factoring.

1. x2 = –x + 6 2. x2 = 5x + 14

3. 2x2 + x = 3 4. 3x2 – 2 = 5x

5. 4x2 = –8x + 5 6. 6x2 = 5x + 6

7-5

Solving Radical EquationsSolving Radical EquationsALGEBRA 2 LESSON 7-5ALGEBRA 2 LESSON 7-5

1. x2 = –x + 6; x2 + x – 6 = 0Factors of –6 with a sum of 1:3 and –2(x + 3)(x – 2) = 0x + 3 = 0 or x – 2 = 0x = –3 or x = 2

3. 2x2 + x = 3; 2x2 + x – 3 = 0(2x + 3)(x – 1) = 02x + 3 = 0 or x – 1 = 0x = – or x = 1

5. 4x2 = –8x + 5; 4x2 + 8x – 5 = 0(2x – 1)(2x + 5) = 02x – 1 = 0 or 2x + 5 = 0

x = or x = –

2. x2 = 5x + 14; x2 – 5x – 14 = 0Factors of –14 with a sum of –5:–7 and 2(x – 7)(x + 2) = 0x – 7 = 0 or x + 2 = 0x = 7 or x = –2

4. 3x2 – 2 = 5x; 3x2 – 5x – 2 = 0(3x + 1)(x – 2) = 03x + 1 = 0 or x – 2 = 0x = – or x = 2

6. 6x2 = 5x + 6; 6x2 – 5x – 6 = 0(3x + 2)(2x – 3) = 03x + 2 = 0 or 2x – 3 = 0

x = – or x =

32

13

12

52

23

32

Solutions

7-5

–10 + 2x + 1 = –5

Solve –10 + 2x + 1 = –5.


2x + 1 = 5 Isolate the radical.

( 2x + 1 )2 = 52 Square both sides.

2x + 1 = 25 2x = 24 x = 12

Check: –10 + 2x + 1 = –5 –10 + 2(12) + 1 –5 –10 + 25 –5 –10 + 5 –5 –5 = –5

7-5


Solve 3(x + 1) = 24.3

5

x + 1 = 32 Simplify. x = 31

3(x + 1) = 243

5

(x + 1) = 8 Divide each side by 3.3

5

Check: 3(x + 1) = 24 3(31 + 1) 24 3(25) 24 3(2)3 24 24 = 24

3

53

53

5

3

5

5

3((x + 1) ) = 8 Raise both sides to the power.

5

3

53

(x + 1)1 = 8 Multiply the exponents and .5

3

53

35

7-5


An artist wants to make a plastic sphere for a sculpture. The

plastic weighs 0.8 ounce per cubic inch. The maximum weight of the

sphere is to be 80 pounds. The formula for the volume V of a sphere

is V = • r 3, where r is the radius of the sphere. What is the maximum radius the sphere can have?

43

7-5

43

Define: Let r = radius in inches.

Relate: volume of sphere • density of plastic maximum weight

Write: • r 3 • 0.8 80<–

<–


(continued)

The maximum radius is about 2.88 inches.

4 r 3

3 • 0.8 80

7-5

r 3 3 • 80

4 • • 0.8<–

r 3 75<–

Use a calculator.r 2.88<–

<–


Solve x + 2 – 3 = 2x. Check for extraneous solutions.

x + 2 – 3 = 2x

x + 2 = 2x + 3 Isolate the radical.

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

0 = 4x2 + 11x + 7 Combine like terms.

0 = (x + 1)(4x + 7) Factor.

x + 2 = 4x2 + 12x + 9 Simplify.

x + 1 = 0 or 4x + 7 = 0 Factor Theorem

x = –1 or x = – 74

7-5

Check: x + 2 – 3 = 2x x + 2 – 3 = 2x

–1 + 2 – 3 2(–1) + 2 – 3 2

1 – 3 –2 – 3

–2 = –2 – 3

72

–

74

–

14

12

72

–

72

–=– 52 /

74

–


(continued)

7-5

The only solution is –1.


Solve (x + 1) – (9x + 1) = 0. Check for extraneous

solutions.

2

3

1

3

x = 0 or x = 7

(x + 1)2 = 9x + 1

x2 + 2x + 1 = 9x + 1

x2 – 7x = 0

x(x – 7) = 0

(x + 1) – (9x + 1) = 02

3

1

3

(x + 1) = (9x + 1) 2

3

1

3

((x + 1) )3 = ((9x + 1) )32

3

1

3

7-5


(continued)

Check: (x + 1) – (9x + 1) = 0 (x + 1) – (9x + 1) = 0

(0 + 1) – (9(0) + 1) 0 (7 + 1) – (9(7) + 1) 0

(1) – (1) 0 (8) – (64 ) 0

(1) – (12) 0 (8) – (82) 0

1 – 1 = 0 8 – 8 = 0

2

3

1

3

2

3

1

3

2

3

1

3

2

3

1

3

2

3

2

3

2

3

1

3

2

3

1

3

2

3

1

3

2

3

1

3

2

3

2

3

Both 0 and 7 are solutions.

7-5



1. 16

2. 1

3. 22

4. 4

5. 23

6.

7. 3

8. 29, –27

9. 18

10. 78

11. 8

23

12. 0

13. 30.6 ft

14. 4 in.

15. 3

16. 1

17. –3, –4

18. 9

19. 1

20. 1

21. 3

22. –2

23. 1

24. 6

25. 2

26. –2

27. 5

28. –3

29. 5

30. –2

31. s = A; 4 2 m,

or about 5.7 m

32. a. s =

b. about 8.8 in.

c. about 15.2 in.

2 3A3

7-5


33. a. b. c.

d. The graph of each pair consists of two straight lines, one of which is horizontal. They intersect at different points, but these points have the same x-value, about 1.236.

34. 8

35. 4

36. 5

37. 23

38. 1

39. 5.0625

40. 6.5

41. 9, –7

42.

43. 9

44. 2

45. –1, –6

46. 2

47. 7

48. 25

49. 10

50. –1

51.

8116

54

7-5


52. d =

53. Answers may vary. Sample: x – 3 = 3x + 5

54. 1

55. 2

56. 0

57. Plan 1: Use a calculator to evaluate 2 + 2 and store the result as x. Evaluate x + 2 and store the result as x. Continue this procedure about seven times until it becomes clear that the values are approaching 2. Plan 2: The given equation is equivalent to x = 2 + x. Solve this equation to find that x = 2.

v2

6458. a. A counterexample is a = 3, b = –3.

b. A counterexample is a = –5, b = 3.

59. 12

60. 79

61.

62.

63. 27

64. 2

65. 16

66. 125

67. 6 2

68. 8 5

54 1 25

7-5


69. 3 2

70.

71. 128

72. 25

73. 2

74.

75. 7

76. 210

77. 60

78. 1680

79. 24

80. 10

3

13

1 106

81. 21

82. 1

83. 6

84. 7

85. 3, 4

86. 3, 5

87. –5, –4

88. –2, –

89. – , –

90. –2, –

23

13

43

34

7-5


Solve each equation. Check for extraneous solutions.

1. 7 + 2x – 1 = 10

2. 4(x – 9) = 8

3. 2x – 1 = x – 8

4. (4x + 3) = (16x + 44)

5. A circular table is to be made that will have a top covered with material that costs $3.50 per square foot. The covering is to cost no more than $60. What is the maximum radius for the top of the table?

1

3

2

3

1

3

5

17

13

74

–,

54

about 2.34 ft

7-5

Function OperationsFunction Operations



Find the domain and range of each function.

1. {(0, –5), (2, –3), (4, –1)} 2. {(–1, 0), (0, 0), (1, 0)}

3. ƒ(x) = 2x – 12 4. g(x) = x2

Evaluate each function for the given value of x.

5. Let ƒ(x) = 3x + 4. Find ƒ(2).

6. Let g(x) = 2x2 – 3x + 1. Find g(–3).

7-6

Function OperationsFunction OperationsALGEBRA 2 LESSON 7-6ALGEBRA 2 LESSON 7-6

1. {(0, –5), (2, –3), (4, –1)}The domain is {0, 2, 4}. The range is {–5, –3, –1}.

2. {(–1, 0), (0, 0), (1, 0)}The domain is {–1, 0, 1}. The range is {0}.

3. ƒ(x) = 2x – 12The domain is the set of all real numbers. Therange is the set of all real numbers.

4. g(x) = x2

The domain is the set of all real numbers. Therange is the set of all nonnegative real numbers.

5. ƒ(x) = 3x + 4; ƒ(2) = 3(2) + 4 = 6 + 4 = 10

6. g(x) = 2x2 – 3x + 1; g(–3) = 2(–3)2 – 3(–3) + 1 =2(9) + 9 + 1 = 18 + 10 = 28

Solutions

7-6


Let ƒ(x) = –2x + 6 and g(x) = 5x – 7. Find ƒ + g and ƒ – g

and their domains.

(ƒ + g)(x) = ƒ(x) + g(x) (ƒ – g)(x) = ƒ(x) – g(x)

= (–2x + 6) + (5x – 7) = (–2x + 6) – (5x – 7)

= 3x – 1 = –7x + 13

The domains of ƒ + g and ƒ – g are the set of real numbers.

7-6


Let ƒ(x) = x2 + 1 and g(x) = x4 – 1. Find ƒ • g and and

their domains.

ƒg

(ƒ • g)(x) = ƒ(x) • g(x)ƒg

ƒ(x)g(x)(x) =

= (x2 + 1)(x4 – 1)x2 + 1x4 – 1

=

= x6 + x4 – x2 – 1 =x2 + 1

(x2 + 1)(x2 – 1)

=1

x2 – 1

The domains of ƒ and g are the set of real numbers, so the domain of ƒ • g is also the set of real numbers.

The domain of does not include 1 and –1 because g(1) and g(–1) = 0. ƒg

7-6


Let ƒ(x) = x3 and g(x) = x2 + 7. Find (g ° ƒ)(2).

Method 1:

(g ° ƒ)(x) = g(ƒ(x)) = g(x3) = x6 + 7

Method 2:

(g ° ƒ)(x) = g(ƒ(x))

(g ° ƒ)(2) = (2)6 + 7 = 64 + 7 = 71 g(ƒ(2)) = g(23)

= g(8)

= 82 + 7

= 71

7-6


A store offers a 20% discount on all items. You have a coupon

worth $3.

a. Use functions to model discounting an item by 20% and to model applying the coupon.

Let x = the original price.

ƒ(x) = x – 0.2x = 0.8x Cost with 20% discount.

g(x) = x – 3 Cost with a coupon for $3.

b. Use a composition of your two functions to model how much you would pay for an item if the clerk applies the discount first and then the coupon.

(g ° ƒ)(x) = g(ƒ(x)) Applying the discount first.= g(0.8x)

= 0.8x – 3

7-6


(continued)

c. Use a composition of your two functions to model how much you would pay for an item if the clerk applies the coupon first and then the discount.

(ƒ ° g)(x) = ƒ(g(x)) Applying the coupon first.= ƒ(x – 3)

= 0.8(x – 3)

= 0.8x – 2.4

d. How much more is any item if the clerk applies the discount first?

(ƒ ° g)(x) – (g ° ƒ)(x) = (0.8x – 2.4) – (0.8x – 3) = 0.6

Any item will cost $.60 more.

7-6



1. x2 + 3x + 5

2. x2 – 3x – 5

3. –x2 + 3x + 5

4. 3x3 + 5x2

5.

6.

7. x2 + 3x + 5

8. –x2 + 3x + 5

9. x2 – 3x – 5

10. 3x3 + 5x2

3x + 5x2

x2

3x + 5

11.

12.

13. 2x2 + 2x – 4; domain: all real numbers

14. –2x2 + 2; domain: all real numbers

15. 2x2 – 2; domain: all real numbers

16. 2x3 – x2 – 4x + 3; domain: all real numbers

17. 2x + 3; domain: all real numbers except 1

18. ; domain: all real numbers except and 1

19. 27x2, domain: all real numbers; 3, domain: all real numbers except 0

20. 2x + 3; 9, –1

21. x2 + 5; 14, 9

3x + 5x2

x2

3x + 5

1 2x + 3

7-6


22. 8

23. 104

24. 20

25. 16

26. 8

27. 10

28. 12

29. 68

30. 404

31. 1

32. 25

33. –3

34. 9

35. 9.25

36. 0.25

37. 6.25

38. –2.75

39. c2 – 6c + 9

40. c2 – 3

41. a2 + 6a + 9

42. a2 – 3

43. a. ƒ(x) = 0.9xb. g(x) = x – 2000c. $14,200d. $14,400

44. a. (g ° ƒ)(x) = 1.0968x

b. 16.45 pesos

45. x2 – x + 7

46. 6x + 13

47. x2 – 5x – 3

48. –2x2 + 8x + 1

49. –x2 + 5x + 13

50. 2x2 + 2x + 24

51. –3x2 + 2x + 16, domain: all real numbers

52. 3x2 – 12, domain: all real numbers

53. 3x3 + 8x2 – 4x – 16, domain: all real numbers

7-6


54. –9x3 – 24x2 + 12x + 48, domain: all real numbers

55. 3x – 4, domain: all real numbers except –2

56. 15x – 20, domain: all real numbers except –2

57. 7; answers may vary. Sample: First evaluate ƒ(3) since the

expression is (g ° ƒ)(3), and that

means g(ƒ(3)). Then evaluate g(6).

58. 1

59. –4

60. 0

61. 2

62. a. 1963; the area after 2 seconds is about 1963 in.2.

b. 7854 in.2

63. 3x2, 9x2

64. x – 2, x – 2

65. 12x2 + 2, 6x2 + 4

66. x – 3, x – 6

67. –4x – 7, –4x – 28

68. ,

69. Answers may vary. Sample:a. g(x) = 0.12xb. ƒ(x) = 9.50xc. (g ƒ)(x) = 1.14x; your savings will

be $1.14 for each hour you work.

x2 + 52

x2 + 10x + 254

7-6

°


70. a. ƒ(x) and g(x)b. 0, 15, 30; 3, 28, 103c. 3x2 + 9d. 3*A1^2 + 9, 9, 84, 309e. 9x2 + 3f. 9*A1^2 + 3, 3, 228, 903

71. a. P(x) = 5295x – 1000b. $157,850

72. a. g(x) is the bonus earned when x is the amount of sales over $5000. h(x) is the excess of x sales over $5000.

b. (g ° h)(x) because you first need

to find the excess sales over $5000 to calculate the bonus.

73. (ƒ+ g)(x) = ƒ(x) + g(x) Def. of Function Add.

= 3x – 2 + (x2 + 1)Substitution = x2 + 3x – 2 + 1 Comm. Prop.= x2 + 3x – 1 arithmetic

74. (ƒ – g)(x) = ƒ(x) – g(x) def. of function subtraction

= 3x – 2 – (x2 + 1)substitution = 3x – 2 – x2 – 1 Opp. of Sum

Prop. = –x2 + 3x – 2 – 1 Comm. Prop. = –x2 + 3x – 3 arithmetic

75. (ƒ ° g)(x) = ƒ(g(x))def. of comp.

functions= ƒ(x2 + 1) substitution= 3(x2 + 1) – 2 substitution= 3x2 + 3 – 2 Dist. Prop.= 3x2 + 1 arithmetic

7-6


76. a. ƒ(x) = x + 10; g(x) = 1.09xb. Each grade is increased 9%

before adding the 10-point bonus; 91.75.

c. Add the 10-point bonus and then increase the sum by 9%; 92.65.

d. no

77. x7 – x6 – 16x5 + 10x4 + 85x3 – 25x2 – 150x; domain: all real numbers

78. ; domain: all real numbers except 3,

5, and – 5

79. ; domain: all real numbers except 0,

–2, 5, and – 5

x2 + 2xx – 3

x – 3 x2 + 2x

80. x

81.

82.

83. 2

84. 4

85. B

86. F

87. [2] (ƒ • g)(x) means ƒ(x) times g(x) or in this case (3x – 4)(x + 3). The value of (ƒ • g)(x) is 3x2 + 5x – 12.

[1] only 3x2 + 5x – 12 OR minor error in explanation

88. A

7-6

1x

6 – x8


89. C

90. B

91. D

92. 1

93. –3

94. 4

95. 3

96. 2

97. 3

98. x8 + 32x7 + 448x6 + 3584x5 + 17,920x4 + 57,344x3 + 114,688x2 + 131,072x + 65,536

99. x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + y6

100. 16x4 – 32x3y + 24x2y2 – 8xy3 + y4

101. 128x7 – 1344x6y + 6048x5y2 – 15,120x4y3 +

22,680x3y4 – 20,412x2y5 + 10,206xy6 – 2187y7

102. 59,049 – 65,610x + 29,160x2 – 6480x3 + 720x4 – 32x5

103. 1024x5 – 1280x4y + 640x3y2 – 160x2y3 + 20xy4 – y5

104. x8 + 4x7 + 6x6 + 4x5 + x4

105. x12 + 12x10y3 + 60x8y6 + 160x6y9 + 240x4y12 + 192x2y15 + 64y18

106. 6 + 2i

107. –2 + 19i 2

108. 13 + 5i 5

109. –8 – 36i

7-6


1. ƒ(x) + g(x)

2. ƒ(x) – g(x)

3. ƒ(x) • g(x)

4.

5. (ƒ ° g)(x)

6. A store is offering a 15% discount on all items. You have a coupon worth $2 off any item. Let x be the original cost of an item. Use a composition of functions to find a function c(x) that gives the final cost of an item if the discount is applied first and then the coupon. Then use this function to find the final cost of an item originally priced at $10.

Let ƒ(x) = 2x2 and g(x) = 3x – 1. Perform each function operation. Then

find the domain.

ƒ(x)g(x)

ƒ(x) + g(x) = 2x2 + 3x – 1; all reals

ƒ(x) – g(x) = 2x2 – 3x + 1; all reals

ƒ(x) • g(x) = 6x3 – 2x2; all reals

=ƒ(x)g(x)

2x2

3x – 1 all reals except13

(ƒ ° g)(x) = 18x2 – 12x + 2; all reals

c(x) = 0.85x – 2; $6.50

7-6

Inverse Relations and FunctionsInverse Relations and Functions



Graph each pair of functions on a single coordinate plane.

1. y = x – 6 2. y = 3. y = 3x – 1

y = x + 6 y = 2x + 7 y =

4. y = 0.5x + 1 5. y = –x + 4 6. y =

y = 2x – 2 y = y = 5x – 4

x – 72

x + 13

–x + 4–1

x + 45

7-7

Inverse Relations and FunctionsInverse Relations and FunctionsALGEBRA 2 LESSON 7-7ALGEBRA 2 LESSON 7-7

1. y = x – 6 2. y = , or

y = x + 6 y = x –

y = 2x + 7

3. y = 3x – 1 4. y = 0.5x + 1

y = , or y = 2x – 2

y = x +

x – 72

12

72

x + 13

13

13

Solutions

7-7


5. y = –x + 4 6. y = , or

y = , or y = x +

y = x – 4 y = 5x – 4

–x + 4–1

x + 45

15

45

Solutions (continued)

7-7


a. Find the inverse of relation m.

Relation m

x –1 0 1 2y –2 –1 –1 –2

Interchange the x and y columns.

Inverse of Relation m

x –2 –1 –1 –2y –1 0 1 2

7-7


(continued)

b. Graph m and its inverse on the same graph.

Relation mReversing theOrdered Pairs Inverse of m

7-7


Find the inverse of y = x2 – 2.

y = x2 – 2

x = y2 – 2 Interchange x and y.

x + 2 = y2 Solve for y.

± x + 2 = y Find the square root of each side.

7-7

The graph of y = –x2 – 2 is a parabola that opens downward with vertex (0, –2).


Graph y = –x2 – 2 and its inverse.

You can also find points on the graph of the inverse by reversing the coordinates of points on y = –x2 – 2.

The reflection of the parabola in the line x = y is the graph of the inverse.

7-7

Consider the function ƒ(x) = 2x + 2 .


a. Find the domain and range of ƒ.

Since the radicand cannot be negative, the domain is the set of numbers greater than or equal to –1.

Since the principal square root is nonnegative, the range is the set of nonnegative numbers.

b. Find ƒ –1

So, ƒ –1(x) = .x2 – 2

2

ƒ(x) = 2x + 2

y = 2x + 2 Rewrite the equation using y.

x = 2y + 2 Interchange x and y.

x2 = 2y + 2 Square both sides.

y =x2 – 2

2 Solve for y.

7-7

(continued)


c. Find the domain and range of ƒ –1.

The domain of ƒ –1 equals the range of ƒ, which is the set of nonnegative numbers.

d. Is ƒ –1 a function? Explain.

For each x in the domain of ƒ–1, there is only one value of ƒ –1(x). So ƒ –1 is a function.

7-7

Note that the range of ƒ–1 is the same as the domain of ƒ.

Since x2 0, –1. Thus the range of ƒ–1 is the set of numbers

greater than or equal to –1.

x2 – 22

>– >–


The function d = 16t 2 models the distance d in feet that an

object falls in t seconds. Find the inverse function. Use the inverse to

estimate the time it takes an object to fall 50 feet.

d = 16t 2

t 2 = d

16Solve for t. Do not interchange variables.

t =d4 Quantity of time must be positive.

t =14 50 1.77

The time the object falls is 1.77 seconds.

7-7

and (ƒ ° ƒ –1)(– 86) = – 86.


For the function ƒ(x) = x + 5, find (ƒ–1 ° ƒ)(652) and

(ƒ ° ƒ–1)(– 86).

12

Since ƒ is a linear function, so is ƒ –1.

Therefore ƒ –1 is a function.

So (ƒ –1 ° ƒ)(652) = 652

7-7



1.

2.

3.

4.

5. y = x – ; yes

6. y = x + ; yes

7. y = – x + ; yes

8. y = y = ± no

9. y = ± x – 4; no

10. y = ± no

11. y = ± x – 1; no

12. y = ± no

13. y = ± ; no

13

13

12

12

13

43

5 – x2

x + 53

x + 42

x – 5 – 12

7-7


14.

15.

16.

17.

18.

19.

20.

21.

22.

7-7


23. ƒ–1(x) = , and the domain

and range for both ƒ and ƒ–1 are all real numbers; ƒ–1 is a function.

24. ƒ–1(x) = x2 + 5, x 0, domain ƒ: {x|x > 5}, range ƒ: {y|y 0}, domain ƒ–1: {x|x 0}, and range ƒ–1: {y|y 5}; ƒ–1 is a function.

25. ƒ–1(x) = x2 – 7, x 0, domain ƒ: {x|x –7}, range ƒ: {y|y 0}, domain ƒ–1: {x|x 0}, and range ƒ–1: {y|y –7}; ƒ–1 is a function.

x – 43

26. ƒ–1(x) = , x 0, domain ƒ: {x|x },

range ƒ: {y|y 0}, domain ƒ–1: {x|x 0}, and range ƒ–1: {y|y }; ƒ–1 is a function.

27. ƒ–1(x) = ± , x 2, domain ƒ: all reals,

range ƒ: {y|y > 2}, domain ƒ–1: {x|x 2}, and range ƒ–1: all reals; ƒ–1 is not a function.

28. ƒ–1(x) = ± 1 – x , x 1, domain ƒ: all reals, range ƒ: {y|y 1}, domain ƒ–1: {x|x 1}, and range ƒ–1: all reals; ƒ–1 is not a function.

3 – x2

3

x – 22

7-7

32

32

>–>–

>–>–

>–>– >–

>–>–

>–

>– >–

>–

>–

<–

<–

<–<– <–


7-7

3

29. a. F = (C – 32); yes

b. –3.89°F

30. a. r = ; yes

b. 20.29 ft

31. 10

32. –10

33. 0.2

34. d

35. ƒ–1(x) = ± ; no

36. ƒ–1(x) = ± 2 ; no

37. ƒ–1(x) = ± ; yes

3V4

3

2x + 83x3

x2 – 6x + 102

38. ƒ–1(x) = ± x – 1; no

39. ƒ–1(x) = ; no

40. ƒ–1(x) = –1 ± x + 1; no

41. ƒ–1(x) = x; yes

42. ƒ–1(x) = ± x; no

43. ƒ–1(x) = ± ; no

44. x = ; 25 ft, 6.25 ft

45. The range of the inverse is the domain of ƒ, which is x 1.

46. 2 and 5

5x – 52

V 2

64

1± x2

4

>–


47. ƒ–1(x) = x2, x 0, domain ofƒ: {x|x 0}, range of ƒ: {y|y 0}, domain of ƒ–1: {x|x 0}, range of ƒ–1: {y|y 0}, and ƒ–1 is a function.

48. ƒ–1(x) = (x – 3)2, x 3, domain of ƒ: {x|x 0}, range of ƒ: {y|y 3}, domain of ƒ–1: {x|x 3}, range of ƒ–1: {y|y 0}, and ƒ–1 is a function.

49. ƒ–1(x) = 3 – x2, x 0, domain of ƒ: {x|x 3}, range of ƒ: {y|y 0}, domain of ƒ–1: {x|x 0}, range of ƒ–1: {y|y 3}, and ƒ–1 is a function.

50. ƒ–1(x) = x2 – 2, x 0, domain of ƒ: {x|x –2}, range of ƒ: {y|y 0}, domain of ƒ–1: {x|x 0}, range of ƒ–1: {y|y –2}, and ƒ–1 is a function.

51. ƒ–1(x) = ± 2x , x 0, domain of ƒ: all reals, range of ƒ: {y|y 0}, domain of ƒ–1: {x|x 0}, range of ƒ–1: all reals, and ƒ–1 is not a function.

52. ƒ–1(x) = ± , x > 0, domain of

ƒ:{x|x = 0}, range of ƒ:{y|y > 0}, domain of ƒ–1: {x|x > 0}, range of ƒ–1: {y|y = 0}, and ƒ–1 is not a function.

/

/

1 x

7-7

>– >–>–

>–

>–>–

>–>–

>–

>–>–

>–

>–>–

>–>–

>–

>–>–

>–

<–

<–

<–


53. ƒ–1(x) = ± x + 4 x 0, domain of ƒ: all reals, range of ƒ: {y|y 0}, domain of ƒ–1: {x|x 0}, range of ƒ–1: all reals, and ƒ–1 is not a function.

54. ƒ–1(x) = 7 ± x x 0, domain of ƒ: all reals, range of ƒ: {y|y 0}, domain of ƒ–1: {x|x 0}, range of ƒ–1: all reals, and ƒ–1 is not a function.

55. ƒ–1(x) = ± – 1 x > 0, domain of

ƒ: {x|x = –1}, range of ƒ: {y|y > 0}, domain of ƒ–1: {x|x > 0}, range of ƒ–1: {y|y = –1}, and ƒ–1 is not a function.

1x

/

/

56. ƒ–1(x) = – x 4, domain of

ƒ: {x|x 0}, range of ƒ: {y|y 4}, domain of ƒ–1: {x 4}, range of ƒ–1: { y|y 0}, and ƒ–1 is a function.

57. ƒ–1(x) = x 0, domain of

ƒ: {x|x > 0}, range of ƒ: {y|y > 0}, domain of ƒ–1: {x|x > 0},range of ƒ–1: {y|y > 0}, and ƒ–1 is a function.

58. ƒ–1(x) = – x > 0, domain of

ƒ: {x|x < 0}, range of ƒ: {y|y > 0}, domain of ƒ–1: {x|x > 0}, range of ƒ–1: {y|y < 0}, and ƒ–1 is a function.

x – 42

12

1x

2

3x

2

7-7

2<–

<–<–

<–

<–<–

>–>–

>–

>–

>–

>–


7-7

59. a-b. Answers may vary. Sample:a.

b.

60. r is not a function because there are two y-values for one x-value. r –1 is a function because each of its x-values has one y-value.

61. h = s 2; 3 2 in. 4.2 in.


63. ƒ–1(x) = 5x; yes

64. ƒ–1(x) = x3 + 5; yes

3

65. ƒ–1(x) = 27x3; yes

66. ƒ–1(x) = 2 + x; yes

67. ƒ–1(x) = x4; yes

68. ƒ–1(x) = ± no

69. B

70. G

71. C

72. [2] x = 4y2 + 5, so 4y2 = x – 5,

y2 = , and y = ± .

The inverse has values that are real numbers when x 5.

[1] y = ± OR x 5 OR minor error

5x6

3

4

x – 54

x – 52

x – 52

>–>–


7-7

73. [4] x = y2 – 2y + 1 or x = (y – 1)2. Then y – 1 = ± x or y = ± x + 1. It is not a function because eachpositive value of x gives two values of y.

[3] minor error in finding inverse[2] attempt to find inverse and a

statement that the inverse is not a function

[1] attempt to find inverse OR a statement that the inverse is not a function

74. 2x + 28

75. 2x + 7

76. |–x – 10|

77. x + 14

78. |–4x – 10|

79. 2x + 28 + |–2x + 4|

80. 2

81. –2

82. not a real number

83. 3

84. –3

85. –3

86. 0.4

87. 30

88. ±1, ±2, ±3, ±4, ±6, ±12, ± , ± ;

roots are – , ±2.

12

32

32


7-7

89. ±1, ±2, ±4, ± , ± , ± ; roots are , 2, –1.

90. ±1, ±2, ±3, ±4, ±6, ±12, ± , ± , ± ; roots are 1, – , -3.

91. ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30, ± , ± , ± , ± ; roots are 5, – , 2.

92. ±1, ±2, ±3, ±6; roots are 1, 2, 3.

93. ±1, ±2, ±3, ±4, ±6, ±12; roots are –2, 2, –3.

13

23

43

23

13

23

43

43

12

32

52

152

32

1. Find the inverse of the function y = 4x – 7. Is the inverse a function?

2. Find the inverse of y = x + 9. Is the inverse a function?

3. Graph the relation y = 2x2 – 2 and its inverse.

4. For the function ƒ(x) = 2x2 – 1, find ƒ–1, and the domain and range of ƒ and ƒ–1. Determine whether ƒ–1 is a function.

5. If ƒ(x) = 5x + 8, find (ƒ–1 °

ƒ)(59) and (ƒ ° ƒ–1)(3001).

6. A right triangle has a leg of length x and a hypotenuse of length 2x. Write an equation to find the length s of the other leg, and use it to estimate s when the hypotenuse has a length of 5 cm.


7-7


1. Find the inverse of the function y = 4x – 7. Is the inverse a function?

2. Find the inverse of y = x + 9. Is the inverse a function?

3. Graph the relation y = 2x2 – 2 and its inverse.

y =x + 7

4 ; yes

y = x2 – 9; yes

7-7


4. For the function ƒ(x) = 2x2 – 1, find ƒ–1, and the domain and range of ƒ and ƒ–1. Determine whether ƒ–1 is a function.

5. If ƒ(x) = 5x + 8, find (ƒ–1 °

ƒ)(59) and (ƒ ° ƒ–1)(3001).

6. A right triangle has a leg of length x and a hypotenuse of length 2x. Write an equation to find the length s of the other leg, and use it to estimate s when the hypotenuse has a length of 5 cm.

ƒ–1(x) = ± 2x + 2; domain and range of ƒ: all numbers, range of ƒ:

all numbers greater than or equal to –1, domain of ƒ–1: all numbers

greater than or equal to –1, range of ƒ–1: all numbers; not a function

12

59, 3001

s = x 3; about 4.33 cm

7-7

Graphing Radical FunctionsGraphing Radical Functions



Graph each equation.

1. y = (x + 2)2 2. y = (x – 3)2 3. y = –(x + 4)2

4. y = –x2 – 1 5. y = –(x + 1)2 + 1 6. y = 3x2 + 3

7-8

Graphing Radical FunctionsGraphing Radical FunctionsALGEBRA 2 LESSON 7-8ALGEBRA 2 LESSON 7-8

1. y = (x + 2)2

parent function: y = x2

translate 2 units left

2. y = (x – 3)2

parent function: y = x2

translate 3 units right

3. y = –(x + 4)2

parent function: y = –x2

translate 4 units left

Solutions

7-8


Solutions (continued)

4. y = –x2 – 1parent function: y = –x2

translate 1 unit down

5. y = –(x + 1)2 + 1parent function: y = –x2

translate 1 unit leftand 1 unit up

6. y = 3x2 + 3parent function: y = 3x2

translate 3 units up

7-8

The graph of y = x + 5 is the graph of

y = x shifted up 5 units.


The domains of both functions are the set of nonnegative numbers, but their ranges differ.

Graph y = x + 5 and y = x – 7.

7-8

The graph y = x – 7 is the graph of y = x shifted down 7 units.

The graph of y = x + 7 is the graph of

y = x shifted left 7 units.

The graph y = x – 5 is the graph of y = x

shifted right 5 units.


Graph y = x + 7 and y = x – 5.

The ranges of both functions are the set of nonnegative numbers, but their domains differ.

7-8


Graph y = – x, y = –2 x, and y = –1.5 x.

13Each y-value of y = –2 x is times the

corresponding y-value of y = –1.5 x.

The domains of the three functions are the set of nonnegative numbers, and the ranges are the set of negative numbers.

7-8

y = 3 x – (–2) + (–1),

so translate the graph of y = 3 x left 2 units

and down 1 unit.


Graph y = 3 x + 2 – 1.

7-8

Graph the model with a graphing calculator.


The function h(x) = 0.5 x models the height h in meters of

one group of male giraffes with a body mass of x kilograms.

3

Use the graph to estimate the body mass of a young male giraffe with a height of 3 meters.

The giraffe has a mass of about 216 kg.

3Graph y = 0.5 x and y = 3. Adjust the window so the graphs intersect.

Use the Intersect feature to find that x 216 when y = 3.

7-8

3

3

The graph of y = 2 x + 2 – 2 is the graph of

y = 2 x translated 2 units left and 2 units down.


Graph y = 2 x + 2 – 2.

7-8

3


Rewrite y = 9x + 18 to make it easy to graph using a

translation. Describe the graph.

y = 9x + 18

= 3 x + 2

= 3 x – (–2)

The graph of y = 9x + 18 is the graph of y = 3 x translated 2 units left.

= 9(x + 2)

7-8



1.

2.

3.

4.

5.

6.

7.

8.

9.

7-8


10.

11.

12.

13.

14.

15.

16.

17.

18.

7-8


19.

20.

21.

22.

23. a.

b. 745 ft, 1053 ft, 1343 ft24.

25.

26.

27.

7-8


28.

29.

30. y = 3 x – 1; the graph is the graph of y = 3 x translated 1 unit to the right.

31. y = –4 x + 2; the graph is the graph of y = –4 x translated 2 units to the left.

32. y = –14 x + 1; the graph is the graph of y = –14 x translated 1 unit to the left.

33. y = 4 x + 2; the graph is the graph of y = 4 x translated 2 units to the left.

34. y = 8 x – 2 – 3; the graph is the graph of y = 8 x translated 2 units to the right and 3 units down.

35. y = 3 x – 2 + 1; the graph is the graph of y = 3 x translated 2 units to the right and 1 unit up.

3

3

3

3

7-8


7-8

36.

D: x 0, R: y 7

37.

D: x 0, R: y –6

38.

D: x 6, R: y 0

39.

D: x 0, R: y 2

40.

D: x 0, R: y 0

41.

D: x , R: y 712

>– >–

>– >–

>– >–

>–

>–

>–

<–

<–<–


7-8

42.

D: all real numbers,R: all real numbers

43.

D: x 1, R: y 3

44.

D: all real numbers, R: all real numbers

45.

D: x – , R: y 012

46.


47.


>–>–

>–<–

50.

D: x , R: y 7

51. a.

b. 4.3 s; 6.1 s


48.

D: x –5, R: y –1

49.

D: all real numbers,R: all real numbers

34

52. a.

b. D: x 2, R: y –2c. (2, –2)d. The domain is based on

the x-coordinate of that point, and the range is based on the y-coordinate.

7-8

>–>–

>– >–<–

<–


7-8

56. y = 6 x + 3 + 4; the graph is the graph of y = 6 x translated 3 units to the left and 4 up.

57. y = – 2 x – ; the graph is the graph of y = –2 x translated unit to the right.

58. y = x – 1 – 2; the graph is the same

as y = x translated 1 unit right and

2 down.

59. y = 10 – x + 3; the graph is the

same as y = – x translated 3 units

to the left and 10 up.

13

13

12

12

14

3

3

3

3

14

>–

>–<–

53. a. y = x – 5 – 2

b. y = x – 1 – 5

54. a.

b. Both domains are x 2. The range of y = x – 2 + 1 is y 1. The range of y = – x – 2 + 1 is y 1.

55. y = 5 x – 4 – 1; the graph is

the same as y = 5 x, translated

4 units to the right and 1 down.


7-8

3

60. y = x + 9 + 5; the graph is the same

as y = x, translated 9 units to the

left and 5 up.

61. Answers may vary. Sample: y = x – 2 + 4

62. a.

b. 20 in.

63. If a > 0, the graph is stretched vertically by a factor of a. If a < 0, the graph is reflected over the x-axis and stretched vertically by a factor of |a|.

13 1

3

64. y = – 2 x + 4; the graph is the graph of y = – 2x translated 4 units to the left; domain: x –4, range: y 0.

65. y = – 8 x – ; the graph is the

graph of y = – 8x translated

units to the right; domain: x ,

range: y 0.

66. y = 3 • x – + 6; the graph is

the graph of y = 3x translated

units to the right and 6 units up;

domain: x , range: y 6.

34

3434

53 5

3

53

>–

>–

>– >–

<–

<–


67. y = – 12 • x – – 3; the graph

is the graph of y = – 12x translated

units to the left and 3 units down;

domain: x – , range: y –3.

68. a.

b. The graph of y = h – x is a reflection of the graph of y = x – h in the line x = h.

69. for all odd positive integers

70. A

71. I

32

32

32

72. B

73. H

74. [2] Both graphs have the same shape as the graph of y = x. The graph of ƒ(x) = x – 1 is the graph of y = x moved 1 unit to the right, and the graph of g(x) = x – 1 is the graph of y = x moved 1 unit down.

[1] minor error in either direction

7-8

<–>–


7-8

81. ƒ–1(x) = ; yes

82. x 3

83.

84.

85.

86.

87.

88.

89. –5 ± 14

90. 6 ± 11

91. , –

48x3y4

2y

3

9 ± 212

–3 ± 3 52

75. [4] For ƒ(x) = x – 1 the domain is x 1 and the range is y 0. For g(x) = x – 1 the domain is x 0 and the range is y –1.

[3] minor error in one of the descriptions of domains or ranges

[2] correct domains or ranges[1] some attempt to describe the domains

76. ƒ–1(x) = ; yes

77. ƒ–1(x) = ; yes

78. ƒ–1(x) = ± ; no

79. ƒ–1(x) = (x + 4)2 – 3; yes

80. ƒ–1(x) = ; no

x + 14

x – 12.4

–1 ± x2

3(x + 3)2

x2

3

–1 ± 6110

12xy2

2y

5

3 9xy2

3y

54

32

>– >–

>–>–

L9.8


23

3

About 2.46 s, about 3.17 s

y = 6 x – 3

7-8

Graph each function on the same graph.

1. y = x – 6 2. y = x + 8

3. y = x 4. y = x – 2 + 5

5. The formula t = 2 can be used to estimate the number ofseconds t it takes a pendulum of length L meters to make one complete swing. Graph the equation on a graphing calculator. Then use the graph to estimate the values of t for pendulums of lengths 1.5 meters and 2.5 meters.

6. Rewrite y = 2 9x – 27 to make it easy to graph using a translation.

Radical Functions and Radical ExponentsRadical Functions and Radical ExponentsALGEBRA CHAPTER 7ALGEBRA CHAPTER 7

7-A

23

x xy3y

6x – 3x 26x

5

1

2

1 25

34x4 x2

9y 6

24. x – 1, domain {x|x = 2}

25. –x3 + 5x2 – 8x + 4, domain: all reals

26. 16x2 + 8x – 1, 4x2 – 7

27. 18x2 + 9x – 6, –6x2 – 3x + 20

28. The sixth power of a real number is always nonnegative.

29. a. ƒ(x) = 0.5xb. g(x) = 0.75xc. g(g(x)) = 0.5625xd. The cashier’s solution is

too high by 6.25% of the original price.

/Page 418

1. –0.3

2. 3|x|y2 6xy

3. –2x2y4 2x4

4. (x – 2)2

5. 7x2 2

6. 6x3y4 2y

7.

8.

9. 24 3

10. 12 2 – 5

11. –17 – 4 5

12. 36 + 38 3

13. – 1 – 6

14. 3 + 2 + 3 + 6

15.

16. x

17.

18. 7

19. 2

20. –43

21. 8

22. x2 – 4x + 4; domain: all reals

23. –2x2 + 7x – 6; domain: all reals


30. y = 4 x + 5 – 1; y = 4 x translated 5 units left and 1 unit down

31. y = 3 x + ; y = 3 x

translated unit left

32.

D: {x|x 0},R: {y|y 3}

33.

D: {x|x – },

R: {y|y 0}

34.

D: {x|x 4},R: {y|y 0}

1313

35.

D: {x|x – },

R: {y|y –4}

36. –5831

37. 192

38. –63

32

32

7-A

>–>–

>–>–

>–

>–>–

<–

4


39. ƒ–1(x) = ; yes

40. g–1(x) = (x + 1)2 – 3; yes

41. g–1(x) = ; yes

42. ƒ–1(x) = ± 4x; no

43. a. V = 268.08

b. r =

c. r 2.88 in.


45. 0.6 seconds; 0.9 seconds

x + 23

3

x2 – 12

3V4

256 3

1

3

7-A

Roots and Radical Expressions

Documents

Transcript of Roots and Radical Expressions