White Plains Middle School€¦ · 238 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved. Chapter 5 35. x 3 = 125 36. 5x = 1080 x = 5 x3 = 216

Copyright © Big Ideas Learning, LLC Algebra 2 235All rights reserved. Worked-Out Solutions

Chapter 5

Chapter 5 Maintaining Mathematical Profi ciency (p. 235)

1. y6 ⋅ y = y6 + 1 = y7 2. n4 —

n3 = n4 − 3= n1 = n

3. x5 —

x6 ⋅ x2 =

x5

— x6 + 2

4. x6 —

x5 ⋅ 3x2 = x6 − 5 ⋅ 3x2

= x5

— x8

= x ⋅ 3x2

= x5 − 8 = 3x2 + 1

= x−3 = 3x3

= 1 —

x3

5. ( 4w3 —

2z2 )

3

= 43(w3)3

— 23(z2)3

= 64w3 ⋅ 3

— 8z2 ⋅ 3

= 8w9

— z6

6. ( m7 ⋅ m — z2 ⋅ m3

) 2

= ( m7 + 1

— z2 ⋅ m3

) 2

= ( m8

— z2 ⋅ m3

) 2

= ( m8 − 3

— z2

) 2

= ( m5

— z2

) 2

= (m5)2

— (z2)2

= m5 ⋅ 2

— z2 ⋅ 2

= m10

— z4

7. 4x + y = 2 8. x − 1 —

3 y = −1

4x − 4x + y = −4x + 2 x − x − 1 —

3 y = −x − 1

y = −4x + 2 − 1 — 3 y = −x − 1

−3 ( − 1 — 3 y ) = −3(−x − 1)

y = 3x + 3

9. 2y − 9 = 13 10. 2xy + 6y = 10

2y − 9 + 9 = 13 + 9 (2x + 6)y = 10

2y = 22 y = 10 —

2x + 6

2y — 2 =

22 —

2 y =

5 —

x + 3

y = 11

11. 8x − 4xy = 3

8x − 8x − 4xy = −8x + 3

−4xy = −8x + 3

−4xy — −4x

= −8x + 3

— −4x

y = −8x + 3

— −4x

= 8x − 3

— 4x

12. 6x + 7xy = 15

6x − 6x + 7xy = −6x + 15

7xy = −6x + 15

7xy — 7x

= −6x + 15

— 7x

y = −6x + 15

— 7x

13. yes; Sample answer: When simplifying x3 ⋅ (x2)2, you must

fi rst apply the Power of a Power Property and then apply the

Product of Powers Property.

Chapter 5 Mathematical Practices (p. 236)

1. Using the Pythagorean Theorem, the equation related to the

smallest triangle is 12 + 12 = a2. Solve the equation for a.

12 + 12 = a2

1 + 1 = a2

2 = a2

a = ± √—

2

Reject the negative solution because length cannot be

negative, so a = √—

2 inches. Repeat this process to fi nd b.

Using the Pythagorean Theorem, the equation related to the

second triangle is 12 + a2 = 12 + ( √—

2 ) 2 = b2. Solve the

equation for b.

12 + ( √—

2 ) 2 = b2

1 + 2 = b2

3 = b2

b = ± √—

3


negative, so b = √—

3 inches. Repeat this process to fi nd

c. Using the Pythagorean Theorem, the equation related to

the third triangle is 12 + b2 = 12 + ( √—

3 ) 2 = c2. Solve the

equation for c.

12 + ( √—

3 ) 2 = c2

1 + 3 = c2

4 = c2

c = ± √—

4

c = ±2


negative, so c = 2 inches. Repeat this process to fi nd

d. Using the Pythagorean Theorem, the equation related

to the last triangle is 12 + c2 = 12 + 22 = d2. Solve the

equation for d.

12 + 22 = d2

1 + 4 = d2

5 = d2

d = ± √—

5


negative, so d = √—

5 inches.

236 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.

Chapter 5

2. From Question 1, the lengths of the sides are a ≈ 1.4 inches,

b ≈ 1.7 inches, c = 2 inches, and d ≈ 2.2 inches.

3. Sample answer: The measurements from a ruler

are a ≈ 1 7 —

16 inches, b ≈ 1

3 —

4 inches, c = 2 inches, and

d ≈ 2 1 —

4 inches.

5.1 Explorations (p. 237)

1. a. √—

9 = 3 = 91/2 b. √—

2 ≈ 1.4 ≈ 21/2

c. 3 √—

8 = 2 = 81/3 d. 3 √—

3 ≈ 1.4 ≈ 31/3

e. 4 √—

16 = 2 = 161/4 f. 4 √—

12 ≈ 1.9 ≈ 121/4

2. a. ( √—

5 ) 3 = ( 51/2 ) 3 b. ( 4 √—

4 ) 2 = ( 41/4 ) 2

= 53/2 = 42/4

≈ 11.18 = 41/2

= 2.00

c. ( 3 √—

9 ) 2 = ( 91/3 ) 2 d. ( 5 √—

10 ) 4 = ( 101/5 ) 4

= 92/3 = 104/5

≈ 4.33 ≈ 6.31

e. ( √—

15 ) 3 = ( 151/2 ) 3 f. ( 3 √—

27 ) 4 = ( 271/3 ) 4

= 153/2 = 274/3

≈ 58.09 = 81

3. a. 82/3 = ( 81/3 ) 2 b. 65/2 = ( 61/2 ) 5

= ( 3 √—

8 ) 2 = ( √

— 6 ) 5

= 4.00 ≈ 88.18

c. 123/4 = ( 121/4 ) 3 d. 103/2 = ( 101/2 ) 3

= ( 4 √—

12 ) 3 = ( √

— 10 ) 3

≈ 6.45 ≈ 31.62

e. 163/2 = ( 161/2 ) 3 f. 206/5 = ( 201/5 ) 6

= ( √—

16 ) 3 = ( 5 √—

20 ) 6

= 64.00 ≈ 36.41

4. A rational exponent can be rewritten as a power involving a

radical by taking the denominator as the index of the radical

and the numerator as the exponent that is placed on the

radical.

5. a. 43/2 = ( 41/2 ) 3

= ( √—

4 ) 3

= 23

= 8

The square root of 4 is 2 and 23 is 8.

b. 324/5 = ( 321/5 ) 4

= ( 5 √—

32 ) 4

= 24

= 16

The fi fth root of 32 is 2 and 24 is 16.

c. 6253/4 = ( 6251/4 ) 3

= ( 4 √—

625 ) 3

= 53

= 125

The fourth root of 625 is 5 and 53 is 125.

d. 493/2 = ( 491/2 ) 3

= ( √—

49 ) 3

= 73

= 343

The square root of 49 is 7 and 73 is 343.

e. 1254/3 = ( 1251/3 ) 4

= ( 3 √—

125 ) 4

= 54

= 625

The cube root of 125 is 5 and 54 is 625.

f. 1006/3 = 1002

= 10,000

The exponent reduces to 2 and 1002 is 10,000.

5.1 Monitoring Progress (pp. 238–240)

1. Because n = 4 is even and a = 16 > 0, 16 has two real

fourth roots. Because 24 = 16 and (−2)4 = 16, you can

write ± 4 √—

16 = ±2 or ±161/4 = ±2.

2. Because n = 2 is even and a = −49 < 0, −49 has no real

square roots.

3. Because n = 3 is odd and a = −125 < 0, −125 has one

real cube root. Because (−5)3 = −125, you can write

3 √—

−125 = −5 or (−125)1/3 = −5.

4. Because n = 5 is odd and a = 243 > 0, 243 has one

real fi fth root. Because 35 = 243, you can write

5 √—

243 = 3 or 2431/5 = 3.

5. 45/2 = ( 41/2 ) 5 = 25 = 32 6. 9−1/2 = ( 91/2 ) −1 = 3−1 = 1 —

3

7. 813/4 = ( 811/4 ) 3 = 33 = 27

8. 17/8 = ( 11/8 ) 7 = 17 = 1

9. 62/5 ≈ 2.05 10. 64−2/3 ≈ 0.06

11. ( 4 √—

16 ) 5 = 165/4 = 32 12. ( 3 √

— −30 ) 2 = (−30)2/3 ≈ 9.65

13. 8x3 = 64 14. 1 — 2 x5 = 512

x3 = 8 x5 = 1024

x = 2 x = 4

The solution is x = 2. The solution is x = 4.


Chapter 5

15. (x + 5)4 = 16 16. (x − 2)3 = −14

x + 5 = ±2 x − 2 ≈ −2.41

x = −5 ± 2 x ≈ −0.41

The solutions are The solution is x ≈ −0.41.

x = −7 and x = −3.

17. The useful life is 10 years, so n = 10. The machine

depreciates to $6000, so S = 6000. The original cost is

$50,000, so C = 50,000. So, the annual depreciation rate is

r = 1 − ( S — C

) 1/n

= 1 − ( 6000 —

50,000 )

1/10

= 1 − ( 3 — 25

) 1/10

≈ 0.191.

The annual depreciation rate is about 0.191, or 19.1%.

5.1 Exercises (pp. 241–242)

Vocabulary and Core Concept Check

1. Rewrite the expression as a−s/t = 1 —

( t √—

a ) s . The index of the

radical is t.

2. For an integer n greater than 1, if bn = a, then b is an nth

root of a.

3. For the real fourth roots, if a is positive, then there are two

real fourth roots, ± 4 √—

a , and if a is negative, then there are no

real fourth roots. For fi fth roots, if a is positive or negative,

then there is one real fi fth root, 5 √—

a .

4. The expression ( m √—

a ) −n does not belong because it has a

negative exponent.

( m √—

a ) −n =

1 —

an/m

( a1/n ) m = ( n √—

a ) m = am/n

Monitoring Progress and Modeling with Mathematics

5. Because n = 3 is odd and a = 8 > 0, 8 has one real cube

root. Because 23 = 8, you can write 3 √—

8 = 2 or 81/3 = 2.

6. Because n = 5 is odd and a = −1 < 0, −1 has one real fi fth

root. Because (−1)5 = −1, you can write 5 √—

−1 = −1 or

(−1)1/5 = −1.

7. Because n = 2 is even and a = 0, 0 has one real square root.

Because 02 = 0, or you can write √—

0 = 0 or 01/2 = 0.

8. Because n = 4 is even and a = 256 > 0, 256 has two real

fourth roots. Because 44 = 256 and (−4)4 = 256, you can

write ± 4 √—

256 = ±4 or ±2561/4 = ±4.


real fi fth root. Because (−2)5 = −32, you can write

5 √—

−32 = −2 or (−32)1/5 = −2.

10. Because n = 6 is even and a = −729 < 0, −729 has no real

sixth roots.

11. 641/6 = (26)1/6 = 2 12. 81/3 = (23)1/3 = 2

13. 253/2 = (251/2)3 = 53 = 125

14. 813/4 = (811/4)3 = 33 = 27

15. (−243)1/5 = [ (−3)5 ] 1/5 = −3

16. (−64)4/3 = [ (−64)1/3 ] 4 = (−4)4 = 256

17. 8−2/3 = (81/3)−2 = 2−2 = 1 —

22 =

1 —

4

18. 16−7/4 = (161/4)−7 = 2−7 = 1 —

27 =

1 —

128

19. The cube root was calculated incorrectly.

272/3 = (271/3)2

= 32

= 9

20. The index and exponent were switched.

2564/3 = ( 3 √—

256 ) 4 ≈ 6.354 ≈ 1625.50

21. B; The denominator of the exponent is 3 and the numerator

is 4.

22. D; The denominator of the exponent is 4 and the numerator

is 3.

23. A; The denominator of the exponent is 4 and the exponent is

negative.

24. C; The denominator of the exponent is 4 and expression is

negative.

25. 5 √—

32,768 = 8 26. 7 √—

1695 ≈ 2.89

27. 25−1/3 ≈ 0.34 28. 851/6 ≈ 2.10

29. 20,7364/5 ≈ 2840.40 30. 86−5/6 ≈ 0.02

31. ( 4 √—

187 ) 3 ≈ 50.57 32. ( 5 √—

−8 ) 8 ≈ 27.86

33. The formula for the volume of a sphere with radius r is

V = 4 —

3 πr3. Substitute 216 for V in the equation and solve for r.

216 = 4 —

3 πr3

162 = πr3

51.566 ≈ r3

3.72 ≈ r

So, the radius of the sphere is about 3.72 feet.

34. The formula for the volume of a cylinder with radius r and

height h is V = πr2h. Substitute 1332 for V and 9 for h in the

equation and solve for r.

1332 = πr2 ⋅ 9

148 = πr2

47.11 ≈ r2

6.86 ≈ r

So, the radius of the cylinder is about 6.86 centimeters.


Chapter 5

35. x3 = 125 36. 5x3 = 1080

x = 5 x3 = 216

The solution is x = 5. x = 6

The solution is x = 6.

37. (x + 10)5 = 70 38. (x − 5)4 = 256

x + 10 ≈ 2.33 x − 5 = ±4

x ≈ −7.66 x = 5 ± 4

The solution is The solutions are x = 1

x ≈ −7.66. and x = 9.

39. x5 = −48 40. 7x4 = 56

x ≈ −2.17 x4 = 8

The solution is x ≈ −2.17. x ≈ ±1.68

The solutions are x ≈ −1.68

and x ≈ 1.68.

41. x6 + 36 = 100 42. x3 + 40 = 25

x6 = 64 x3 = −15

x = ±2 x ≈ −2.47


x = −2 and x = 2.

43. 1 — 3 x4 = 27 44. 1 —

6 x3 = −36

x4 = 81 x3 = −216

x = ±3 x = −6

The solutions are The solution is x = −6.

x = −3 and x = 3.

45. The period of time for each case is 100 years, so n = 100.

Potatoes: The price in 1913 was $0.016 per pound, so

p1 = 0.016. The price in 2013 was $0.627 per pound,

so p2 = 0.627. So, the infl ation rate is

r = ( p2 — p1

) 1/n

− 1 = ( 0.627 —

0.016 )

1/100

− 1 ≈ 0.024.

The infl ation rate for potatoes is about 0.024, or 2.4%.

Ham: The price in 1913 was $0.251 per pound, so

p1 = 0.251. The price in 2013 was $2.693 per pound, so

p2 = 2.693. So, the infl ation rate is

r = ( p2 — p1

) 1/n

− 1 = ( 2.693 —

0.251 )

1/100

− 1 ≈ 0.037.

The infl ation rate for ham is about 0.037, or 3.7%.

Eggs: The price in 1913 was $0.373 per dozen, so

p1 = 0.373. The price in 2013 was $1.933 per dozen,

so p2 = 1.933. So, the infl ation rate is

r = ( p2 — p1

) 1/n

− 1 = ( 1.933 —

0.373 )

1/100

− 1 ≈ 0.017.

The infl ation rate for eggs is about 0.017, or 1.7%.

46. From the shape of the graph, n is an even number. The

number a is positive and has two real nth roots because the

graphs intersect at only two points.

47. The number 4 √—

125 lies between 3 and 4 because 125 lies

between 81 = 34 and 256 = 44.

48.

x

y

(1, 0) (1.52, 0)

Earth Mars

(1

Ea

Because it takes Mars 1.88 years to orbit the Sun, substitute

1.88 into the equation for t and solve for d.

d3 = 1.882

d = 3 √—

1.882

d ≈ 1.52

So, the distance Mars is from the Sun is about

1.52 astronomical units.

49. Because the length of the bottom is 20 feet, use ℓ = 20

and because the depth of the water is 5 feet, use h = 5.

So, Q = 3.367(20)(5)3/2 ≈ 753. So, there is about

753 cubic feet of water per second going over the spillway.

50. The equation that relates the particle mass to the speed of

the river is m = ks6, where m is the mass of the particle, s is

the speed of the river, and k is a constant. Because the k is a

constant, it does not change when the speed of the river or

the mass of the particle changes. When the mass is doubled,

multiply each side of the formula by 2 and rewrite.

2m = 2ks6

2m = k ⋅ 2s6

2m = k ( 21/6s ) 6

So, the speed of the river must be 21/6 ≈ 1.12 times faster

for a particle double the mass to be transported, or about

1.12 meters per second. Similarly, for a particle 10 times

the mass, the speed needs to be 101/6 ≈ 1.47 times faster, or

about 1.47 meters per second, and for a particle 100 times

the mass, the speed needs to be 1001/6 ≈ 2.15 times faster,

or about 2.15 meters per second.

Maintaining Mathematical Profi ciency

51. 5 ⋅ 54 = 54 + 1 = 55 52. 42 —

47 = 42 − 7 = 4−5 =

1 —

45

53. (z2)−3 = z2(−3) = z−6 = 1 —

z6

54. ( 3x —

2 )

4

= (3x)4

— 24

= 34x4

— 16

= 81x4

— 16

55. 5 × 103 = 5000 56. 4 × 10−2 = 0.04

57. 8.2 × 10−1 = 0.82 58. 6.93 × 106 = 6,930,000


Chapter 5


1. a. a−2 = 1 —

a2 , a ≠ 0; D b. (ab)4 = a4b4; C

c. (a3)4 = a12; B d. a3 ⋅ a4 = a7; A

e. ( a — b )

3 =

a3

— b3

, b ≠ 0; G f. a6 —

a2 = a4, a ≠ 0; F

g. a0 = 1, a ≠ 0; E

2. a. 52/3 ⋅ 54/3 = 52/3 + 4/3 b. 31/5 ⋅ 34/5 = 31/5 + 4/5

= 56/3 = 35/5

= 52 = 31

= 25 = 3

c. (42/3)3 = 4(2 ⋅ 3)/3 d. (101/2)4 = 104/2

= 42 = 102

= 16 = 100

e. 85/2 —

81/2 = 85/2 − 1/2 f.

72/3

— 75/3

= 72/3 − 5/3

= 84/2 = 7−3/3

= 82 = 7−1

= 64 = 1 —

7

3. a. √—

3 ⋅ √—

12 = 31/2 ⋅ 121/2 = (3 ⋅ 12)1/2 = √—

3 ⋅ 12

√—

3 ⋅ 12 = √—

36 = 6

b. 3 √—

5 ⋅ 3 √—

25 = 51/3 ⋅ 251/3 = (5 ⋅ 25)1/3 = 3 √—

5 ⋅ 25

3 √—

5 ⋅ 25 = 3 √—

125 = 5

c. 4 √—

27 ⋅ 4 √—

3 = 271/4 ⋅ 31/4 = (27 ⋅ 3)1/4 = 4 √—

27 ⋅ 3

4 √—

27 ⋅ 3 = 4 √—

81 = 3

d. √

— 98 —

√—

2 =

981/2

— 21/2

= ( 98 —

2 ) 1/2

= √—

98

— 2

√—

98

— 2 = √

— 49 = 7

e. 4 √—

4 —

4 √—

1024 =

41/4

— 10241/4

= ( 4 —

1024 ) 1/4

= 4 √—

4 —

1024

= 4 √—

4 —

1024 = 4

√—

1 —

256 =

1 —

4

f. 3 √—

625 —

3 √—

5 =

6251/3

— 51/3

= ( 625 —

5 ) 1/3

= 3 √—

625

— 5

3 √—

625

— 5 =

3 √—

125 = 5

4. You can rewrite the radical with a rational exponent, and

then use the properties of exponents to add or subtract

exponents.

5. a. √—

27 ⋅ √—

6 = √—

27 ⋅ 6

= √—

162

= √—

81 ⋅ 2

= √—

81 ⋅ √—

2

= 9 √—

2

b. 3 √—

240 —

3 √—

15 = 3

√—

240

— 15

= 3 √—

16

= 3 √—

8 ⋅ 2

= 3 √—

8 ⋅ 3 √—

2

= 2 3 √—

2

c. (51/2 ⋅ 161/4)2 = ( 51/2 ⋅ 2 ) 2

= (51/2)2 ⋅ 22

= 52/2 ⋅ 4

= 5 ⋅ 4

= 20


1. 23/4 ⋅ 21/2 = 2(3/4 + 1/2) = 25/4

2. 3 —

31/4 =

31

— 31/4

= 3(1 − 1/4) = 33/4

3. ( 201/2 —

51/2 )

3

= [ ( 20 —

5 )

1/2

] 3 = ( 41/2 ) 3 = (2)3 = 8

4. ( 51/3 ⋅ 71/4 ) 3 = ( 51/3 ) 3 ⋅ ( 71/4 ) 3

= 5(1/3) ⋅ 3 ⋅ 7(1/4) ⋅ 3

= 51 ⋅ 73/4

= 5 ⋅ 73/4

5. 4 √—

27 ⋅ 4 √—

3 = 4 √—

27 ⋅ 3 = 4 √—

81 = 3

6. 3 √—

250 —

3 √—

2 = 3

√—

250

— 2 =

3 √—

125 = 5

7. 3 √—

104 = 3 √—

8 ⋅ 13 = 3 √—

8 ⋅ 3 √—

13 = 2 3 √—

13

8. 5 √—

3 —

4 = 5

√—

3 —

4 ⋅ 5

√—

8 —

8

= 5 √—

3 —

4 ⋅

8 —

8 = 5

√—

24

— 32

= 5 √—

24 —

5 √—

32 =

5 √—

24 —

2

9. 3 —

6 − √—

2 =

3 —

6 − √—

2 ⋅

6 + √—

2 —

6 + √—

2

= 3 ( 6 + √

— 2 ) —

62 − ( √—

2 ) 2

= 18 + 3 √

— 2 —

36 − 2

= 18 + 3 √

— 2 —

34


Chapter 5

10. 7 5 √—

12 − 5 √—

12 = (7 − 1) 5 √—

12

= 6 5 √—

12

11. 4 ( 92/3 ) + 8 ( 92/3 ) = (4 + 8) ( 92/3 )

= 12 ( 92/3 )

= 12 [ ( 92 ) 1/3 ]

= 12(811/3)

= 12 [ (27 ⋅ 3)1/3 ]

= 12(271/3 ⋅ 31/3)

= 12(3 ⋅ 31/3)

= 36 ( 31/3 )

= 36 3 √—

3

12. 3 √—

5 + 3 √—

40 = 3 √—

5 + 3 √—

5 ⋅ 8

= 3 √—

5 + 3 √—

5 ⋅ 3 √—

8

= 3 √—

5 + 2 3 √—

5

= (1 + 2) 3 √—

5

= 3 3 √—

5

13. 3 √—

27q9 = 3 √—

33q9 = 3q3 14. 5 √—

x10

— y5

= 5 √—

x10 —

5 √—

y5 =

x2

— y

15. 6xy3/4 —

3x1/2y1/2 = 2x[1 − (1/2)]y[(3/4) − (1/2)] = 2x1/2y1/4

16. √—

9w5 − w √—

w3 = √—

9w4w1 − w √—

w2w

= 3w2 √— w − w2 √—

w

= (3w2 − w2) √— w

= 2w2 √— w



1. A radical expression with index n is in simplest form when

no radicands have perfect nth powers other than 1 as factors,

no radicands contain fractions, and no radicals appear in the

denominator of a fraction.

2. The fi rst radical, 3 √—

4 —

5 , does not belong with the other three

because it is not in simplest form.


3. (92)1/3 = 92/3 4. (122)1/4 = 12[2 ⋅ (1/4)] = 121/2

5. 6 —

61/4 = 61 − (1/4) = 63/4 6. 7

— 71/3

= 7[1 − (1/3)] = 72/3

7. ( 84 —

104 )

−1/4

= [ ( 8 — 10

) 4 ] −1/4

8. ( 93

— 63

) −1/3

= [ ( 9 — 6 )

3

] −1/3

= ( 8 — 10

) 4 ⋅ (−1/4)

= ( 9 — 6 )

3 ⋅ (−1/3)

= ( 8 —

10 ) −1

= ( 9 — 6 ) −1

= 10

— 8 =

5 —

4 = 6 —

9 =

2 —

3

9. ( 3−2/3 ⋅ 31/3 ) −1 = ( 3−2/3 ) −1 ⋅ ( 31/3 ) −1

= 32/3 ⋅ 3−1/3

= 3(2/3) + (−1/3)

= 31/3

OR

( 3−2/3 ⋅ 31/3 ) −1 = ( 3−2/3 + 1/3 ) −1

= ( 3−1/3 ) −1

= 31/3

10. ( 51/2 ⋅ 5−3/2 ) −1/4 = [ 51/2 + (−3/2) ] −1/4

= ( 5−2/2 ) −1/4

= ( 5−1 ) −1/4

= 51/4

11. 22/3 ⋅ 162/3

— 42/3

= ( 2 ⋅ 16 —

4 )

2/3

= 82/3 = (81/3)2 = (2)2 = 4

12. 493/8 ⋅ 497/8 —

75/4 =

(72)3/8 ⋅ (72)7/8

—— 75/4

= 73/4 ⋅ 77/4

— 75/4

= 7(3/4 + 7/4)

— 75/4

= 710/4

— 75/4

= 7(10/4 − 5/4)

= 75/4

13. √—

2 ⋅ √—

72 = √—

2 ⋅ 72 14. 3 √—

16 ⋅ 3 √—

32 = 3 √—

16 ⋅ 32

= √—

144 = 3 √—

512

= 12 = 8

15. 4 √—

6 ⋅ 4 √—

8 = 4 √—

6 ⋅ 8 16. 4 √—

8 ⋅ 4 √—

8 = 4 √—

8 ⋅ 8

= 4 √—

48 = 4 √—

64

= 4 √—

16 ⋅ 3 = 4 √—

16 ⋅ 4

= 2 4 √—

3 = 2 4 √—

4

17. 5 √—

486 —

5 √—

2 = 5

√—

486

— 2 18.

√—

2 —

√—

32 = √—

2 —

32

= 5 √—

243 = 3 = √—

1 —

16 =

1 —

4

19. 3 √—

6 ⋅ 3 √—

72 —

3 √—

2 = 3

√— 6 ⋅ 72

— 2 20.

3 √—

3 ⋅ 3 √—

18 —

6 √—

2 ⋅ 6 √—

2 =

3 √—

3 ⋅ 18 —

6 √—

2 ⋅ 2

= 3 √—

216 = 6 = 3 √—

54 —

6 √—

22

= 3 √—

54 —

3 √—

2

= 3 √—

54

— 2

= 3 √—

27 = 3


Chapter 5

21. 4 √—

567 = 4 √—

81 ⋅ 7 22. 5 √—

288 = 5 √—

32 ⋅ 9

= 4 √—

81 4 √—

7 = 5 √—

32 5 √—

9

= 3 4 √—

7 = 2 5 √—

9

23. 3 √—

5 —

3 √—

4 =

3 √—

5 —

3 √—

4 ⋅

3 √—

2 —

3 √—

2 24.

4 √—

4 —

4 √—

27 =

4 √—

4 —

4 √—

27 ⋅

4 √—

3 —

4 √—

3

= 3 √—

10 —

3 √—

8 =

4 √—

12 —

4 √—

81

= 3 √—

10 —

2 =

4 √—

12 —

3

25. √—

3 —

8 =

√—

3 —

√—

8 ⋅

√—

2 —

√—

2 26. 3

√—

7 —

4 =

3 √—

7 —

3 √—

4 ⋅

3 √—

2 —

3 √—

2

= √

— 6 —

√—

16 =

3 √—

14 —

3 √—

8

= √

— 6 —

4 =

3 √—

14 —

2

27. 3 √—

64

— 49

= 3 √—

64 —

3 √—

49 ⋅

3 √—

7 —

3 √—

7 28. 4

√—

1296

— 25

= 4 √—

1296 —

4 √—

25 ⋅

4 √—

25 —

4 √—

25

= 4

3 √—

7 —

3 √—

343 =

6 4 √—

25 —

4 √—

625

= 4

3 √—

7 —

7 =

6 4 √—

25 —

5

= 6 √

— 5 —

5

29. 1 —

1 + √—

3 =

1 —

1 + √—

3 ⋅

1 − √—

3 —

1 − √—

3

= 1 ( 1 − √

— 3 ) —

12 − ( √—

3 ) 2

= 1 − √

— 3 —

1 − 3

= 1 − √

— 3 —

−2

30. 1 —

2 + √—

5 =

1 —

2 + √—

5 ⋅

2 − √—

5 —

2 − √—

5

= 1 ( 2 − √

— 5 ) —

22 − ( √—

5 ) 2

= 2 − √

— 5 —

4 − 5

= 2 − √

— 5 —

−1

= √—

5 − 2

31. 5 —

3 − √—

2 =

5 —

3 − √—

2 ⋅

3 + √—

2 —

3 + √—

2

= 5 ( 3 + √

— 2 ) —

32 − ( √—

2 ) 2

= 15 + 5 √

— 2 —

9 − 2

= 15 + 5 √

— 2 —

7

32. 11 —

9 − √—

6 =

11 —

9 − √—

6 ⋅

9 + √—

6 —

9 + √—

6

= 11 ( 9 + √

— 6 ) —

92 − ( √—

6 ) 2

= 99 + 11 √

— 6 —

81 − 6

= 99 + 11 √

— 6 —

75

33. 9 —

√—

3 + √—

7 =

9 —

√—

3 + √—

7 ⋅

√—

3 − √—

7 —

√—

3 − √—

7

= 9 ( √

— 3 − √

— 7 ) ——

( √—

3 ) 2 − ( √—

7 ) 2

= 9 √

— 3 − 9 √

— 7 —

3 − 7

= 9 √

— 3 − 9 √

— 7 —

−4

34. 2 —

√—

8 + √—

7 =

2 —

√—

8 + √—

7 ⋅

√—

8 − √—

7 —

√—

8 − √—

7

= 2 ( √

— 8 − √

— 7 ) ——

( √—

8 ) 2 − ( √—

7 ) 2

= 2 √

— 8 − 2 √

— 7 —

8 − 7

= 2 √—

8 − 2 √—

7

= 4 √—

2 − 2 √—

7

35. √

— 6 —

√—

3 − √—

5 =

√—

6 —

√—

3 − √—

5 ⋅

√—

3 + √—

5 —

√—

3 + √—

5

= √

— 6 ( √

— 3 + √

— 5 ) ——

( √—

3 ) 2 − ( √—

5 ) 2

= √

— 18 + √

— 30 —

3 − 5

= √

— 18 + √

— 30 —

−2

= 3 √

— 2 + √

— 30 —

−2


Chapter 5

36. √

— 7 —

√—

10 − √—

2 =

√—

7 —

√—

10 − √—

2 ⋅

√—

10 + √—

2 —

√—

10 + √—

2

= √

— 7 ( √

— 10 + √

— 2 ) ——

( √—

10 ) 2 − ( √—

2 ) 2

= √

— 70 + √

— 14 —

10 − 2

= √

— 70 + √

— 14 —

8

37. 9 3 √—

11 + 3 3 √—

11 = (9 + 3) 3 √—

11

= 12 3 √—

11

38. 8 6 √—

5 − 12 6 √—

5 = (8 − 12) 6 √—

5

= −4 6 √—

5

39. 3(111/4) + 9(111/4) = (3 + 9)(111/4)

= 12(111/4)

40. 13(83/4) − 4(83/4) = (13 − 4)(83/4)

= 9(83/4)

41. 5 √—

12 − 19 √—

3 = 5 √—

4 ⋅ 3 − 19 √—

3

= 5 √—

4 √—

3 − 19 √—

3

= 10 √—

3 − 19 √—

3

= (10 − 19) √—

3

= −9 √—

3

42. 27 √—

6 + 7 √—

150 = 27 √—

6 + 7 √—

6 ⋅ 25

= 27 √—

6 + 7 √—

6 √—

25

= 27 √—

6 + 35 √—

6

= (27 + 35) √—

6

= 62 √—

6

43. 5 √—

224 + 3 5 √—

7 = 5 √—

7 ⋅ 32 + 3 5 √—

7

= 5 √—

32 ⋅ 5 √—

7 + 3 5 √—

7

= 2 5 √—

7 + 3 5 √—

7

= (2 + 3) 5 √—

7

= 5 5 √—

7

44. 7 3 √—

2 − 3 √—

128 = 7 3 √—

2 − 3 √—

64 ⋅ 2

= 7 3 √—

2 − 3 √—

64 3 √—

2

= 7 3 √—

2 − 4 3 √—

2

= (7 − 4) 3 √—

2

= 3 3 √—

2

45. 5(241/3) − 4(31/3) = 5 [ (8 ⋅ 3)1/3 ] − 4(31/3)

= 5(81/3 ⋅ 31/3) − 4(31/3)

= 5 ⋅ 2(31/3) − 4(31/3)

= (10 − 4)(31/3)

= 6(31/3)

46. 51/4 + 6(4051/4) = 51/4 + 6 [ (81 ⋅ 5)1/4 ]

= 51/4 + 6(811/4 ⋅ 51/4)

= 51/4 + 6(3 ⋅ 51/4)

= (1 + 18)(51/4)

= 19(51/4)

47. The radicand should not change when the expression is

factored.

3 3 √—

12 + 5 3 √—

12 = (3 + 5) 3 √—

12 = 8 3 √—

12

48. (A) (52/9)3/2 = 51/3

(B) 53

— ( 3 √

— 5 ) 8

= 53

— 58/3

= 53 − (8/3) = 51/3

(C) 3 √—

625 = 3 √—

54 = 54/3 = 5(51/3)

(D) 3 √—

5145 − 3 √—

875 = 3 √—

5 ⋅ 3 ⋅ 73 − 3 √—

53 ⋅ 7

= 7 ⋅ 31/351/3 − 5 ⋅ 71/3

(E) 3 √—

5 + 3 3 √—

5 = 4 3 √—

5 = 4(51/3)

(F) 7 4 √—

80 − 2 4 √—

405 = 7 4 √—

16 ⋅ 5 − 2 4 √—

81 ⋅ 5

= 14(51/4) − 6(51/4) = 8(51/4)

So, (A), (B), (C), and (E) are like radicals.

49. 4 √—

81y8 = 4 √—

34 ⋅ (y2)4

= 4 √—

34 ⋅ 4 √—

(y2)4

= 3y2

50. 3 √—

64r3t6 = 3 √—

43r3(t2)3

= 3 √—

43 3 √—

r3 3 √—

(t2)3

= 4rt2

51. 5 √—

m10

— n5

= 5 √—

(m2)5

— n5

= 5 √—

(m2)5 —

5 √—

n5

= m2

— n

52. 4 √—

k16

— 16z4

= 4 √—

(k4)4 —

4 √—

24z4

= k4

— 2 ∣ z ∣

53. 6 √—

g6h

— h7

= 6 √— g6 ⋅

h —

h7

= 6 √—

g6

— h6

= 6 √—

( g — h )

6

= ∣ g — h ∣


Chapter 5

54. 8 √— n18p7

— n2p−1

= 8 √——

n(18 − 2) ⋅ p[7 − (−1)]

= 8 √—

n16p8

= 8 √—

(n2)8 8 √—

p8

= n2 ∣ p ∣ , n ≠ 0, p ≠ 0

55. Because 6 is even, to ensure all variables are positive,

absolute value should have been used:

6 √—

g6 = ∣ g ∣

6 √—

64h12

— g6

= 6 √—

64h12 —

6 √—

g6

= 6 √—

26(h2)6 —

6 √—

g6

= 2h2

— ∣ g ∣

56. Sample answer: √—

x2 and √—

x4 are two expressions involving

radicals.

√—

x2 = ∣ x ∣ and √—

x4 = √—

(x2)2 = x2

The fi rst expression needs an absolute value because the

variable x can be positive, negative, or zero.

The second expression does not need an absolute value

because x2 is always positive.

57. √—

81a7b12c9 = √——

92(a3)2a(b6)2(c4)2c

= √——

92(a3)2(b6)2(c4)2 √—

ac

= 9a3b6c4 √—

ac

58. 3 √—

125r4s9t7 = 3 √——

53r3r(s3)3(t2)3t

= 3 √——

53r3(s3)3(t2)3 3 √—

rt

= 5rs3t2 3 √—

rt

59. 5 √— 160m6

— n7

= 5 √—

(32)(5)m5m ——

5 √—

n5n2

= 5 √—

32m5 5 √—

5m ——

5 √—

n5 5 √—

n2

= 2m

5 √—

5m —

n 5 √—

n2 ⋅

5 √—

n3 —

5 √—

n3

= 2m

5 √—

5mn3 —

n 5 √—

n2n3

= 2m

5 √—

5mn3 —

n ⋅ n

= 2m

— n2

5 √—

5mn3

60. 4 √— 405x3y3

— 5x−1y

= 4 √——

81x3 − (−1)y3 − 1 = 4 √—

81x4y2 = 3x √—

y , y ≠ 0

61. 3 √

— w ⋅ √

— w5 —

√—

25w16 =

w1/3 ⋅ w5/2

— 5w8

62. 4 √—

v6 —

7 √—

v5 =

v6/4

— v5/7

= w1/3 + 5/2

— 5w8

= v3/2

— v5/7

= w17/6

— 5w8

= v(3/2 − 5/7)

= w2 ⋅ w5/6

— 5w8

= v11/14

= 6 √—

w5 —

5w6 =

14 √—

v11 , v ≠ 0

63. 18w1/3v5/4 —

27w4/3v1/2 = 2 —

3 w(1/3 − 4/3)v(5/4 − 1/2)

= 2 —

3 w−3/3v3/4

= 2 —

3 w−1v3/4

= 2v3/4

— 3w

, v ≠ 0

64. 7x−3/4y5/2z−2/3 ——

56x−1/2y1/4 =

1 —

8 ⋅

x1/2y5/2 − 1/4

— x3/4z2/3

= 1 —

8 ⋅

x1/2y9/4

— x3/4z2/3

= 1 —

8 ⋅

x1/2 + 1/4y9/4z0 + 1/3

—— x3/4 + 1/4z2/3 + 1/3

= x3/4y9/4z1/3

— 8xz

, y ≠ 0

65. 12 3 √—

y + 9 3 √—

y = (12 + 9) 3 √—

y

= 21 3 √—

y

66. 11 √—

2z − 5 √—

2z = (11 − 5) √—

2z

= 6 √—

2z

67. 3x7/2 − 5x7/2 = (3 − 5)x7/2

= −2x7/2

68. 7 3 √—

m7 + 3m7/3 = 7m7/3 + 3m7/3

= (7 + 3)m7/3

= 10m7/3

69. 4 √—

16w10 + 2w 4 √—

w6 = 4 √—

16 w10/4 + 2w1w6/4

= 2w10/4 + 2w10/4

= (2 + 2)w10/4

= 4w10/4

= 4w5/2

= 4w2 √—

w

70. (p1/2 ⋅ p1/4) − 4 √—

16p3 = p3/4 − 4 √—

16 p3/4

= p3/4 − 2p3/4

= (1 − 2)p3/4

= (−1)p3/4

= −p3/4


Chapter 5

71. P = 2(x3 + 2x2/3) A = x3 ⋅ 2x2/3

= 2x3 + 4x2/3 = 2x[3 + (2/3)]

= 2x11/3

72. P = 3x1/3 + 4x1/3 + √——

(3x1/3)2 + (4x1/3)2

= (3 + 4)x1/3 + √——

9x2/3 + 16x2/3

= 7x1/3 + √—

(9 + 16)x2/3

= 7x1/3 + √—

25x2/3

= 7x1/3 + 5x1/3

= (7 + 5)x1/3

= 12x1/3

A = 1 —

2 ⋅ 4x1/3 ⋅ 3x1/3

= 2 ⋅ 3x(1/3 + 1/3)

= 6x2/3

73. d = 1.9 [ ( 5.5 × 10−4 ) 100 ] 1/2

= 1.9[(0.00055)(100)]1/2

= 1.9(0.055)1/2

≈ 0.45

The optimum pinhole diameter for the camera box is about

0.45 millimeter.

74. a. m = 32 grams

S = (57.5)(32)2/3

= (57.5)(25)2/3

= (57.5)210/3

= (57.5)2321/3

= 460 3 √—

2

≈ 579.56

The surface area of a bat is about 579.56 square centimeters.

b. m = 3.4 kilograms = 3.4 × 103 grams

S = (9.75)(3.4 × 103)2/3

= (9.75)(3.4)2/3(102)

≈ 2204.57

The surface area of a rabbit is about 2204.57 square centimeters.

c. m = 59 kilograms = 59 × 103 grams

S = (11.0)(59 × 103)2/3

= (11)(59)2/3(102)

≈ 16,670.96

The surface area of a human is about 16,670.96 square

centimeters.

75. Your friend is not correct. The second radical can be

simplifi ed to 18 √—

11 .

7 √—

11 − 9 √—

44 = 7 √—

11 − 9 √—

11 ⋅ 4

= 7 √—

11 − 9 √—

11 √—

4

= 7 √—

11 − 18 √—

11

= (7 − 18) √—

11

= −11 √—

11

76. a. m1 = 0.77, m2 = 0.03

2.5120.77

— 2.5120.03

= (2.512)0.77 − 0.03 = (2.512)0.74 ≈ 1.98

Altair is about 1.98 times fainter than Vega.

b. m1 = 1.25, m2 = 0.77

2.5121.25

— 2.5120.77

= (2.512)1.25 − 0.77 = (2.512)0.48 ≈ 1.56

Deneb is about 1.56 times fainter than Altair.

c. m1 = 1.25, m2 = 0.03

2.5121.25

— 2.5120.03

= (2.512)1.25 − 0.03 = (2.512)1.22 ≈ 3.08

Deneb is about 3.08 times fainter than Vega.

77. Perimeter of triangle = √—

82 + 42 + √—

42 + 22

+ √——

82 + (8 − 2)2

= √—

64 + 16 + √—

16 + 4 + √—

64 + 36

= √—

80 + √—

20 + √—

100

= √—

5 ⋅ 16 + √—

5 ⋅ 4 + √—

102

= 4 √—

5 + 2 √—

5 + 10

= 6 √—

5 + 10

78. f(x) = √—

64x2 = ∣ 8x ∣ So, the graph of f is B.

g(x) = 3 √—

64x6 = 4x2

So, the graph of g is A.

79. a. V = 4 —

3 πr3

3V = 4πr3

3V — 4π

= r3

3 √—

3V

— 4π

= r

b. S = 4πr2

= 4π ( 3 √—

3V

— 4π

) 2

= 4π [ ( 3V —

4π ) 1/3

] 2

= 4π ( 3V —

4π )

2/3

= (4π)1

— (4π)2/3

(3V )2/3

= (4π)[1 − (2/3)](3V)2/3

= (4π)1/3(3V)2/3

c. Surface area of one balloon = S1 = (4π)1/3(3V )2/3

Surface area of another balloon = S2 = (4π)1/3 [ 3(2V) ] 2/3

= (4π)1/3(3V )2/3 ⋅ 22/3

So, the surface area of one balloon is 22/3 ≈ 1.59 times the

surface area of the other balloon.


Chapter 5

80. (x2)1/6 = x2/6 = x1/3

(x1/6)2 = x2/6 = x1/3

The expressions are not equivalent for all values of x. When

x is negative, the fi rst expression is always positive and the

second expression is not a real number.

81. Absolute value is needed when n is an even number and m

— n

is an odd number.


82. Step 1 Identify the focus, directrix, and axis of symmetry.

The equation has the form y = 1 —

4p x2, where p =

1 —

8 . The

focus is (0, p) or ( 0, 1 —

8 ) . The directrix is y = −p, or

y = − 1 —

8 . Because x is squared, the axis of symmetry is

the y-axis, x = 0.

Step 2 Use a table of values to graph the equation. Notice

that it is easier to substitute x-values and solve for y.

Opposite x-values result in the same y-value.

x 0 ±1 ±2 ±3 ±4

y 0 2 8 18 32

x

y

4

5

6

7

8

9

10

2

3

1

−2

−1y = −1

8—

(0, )18

x = 0

−3 −2 1

83. Step 1 Rewrite the equation in standard form.

y2 = −x

x = −y2

Step 2 Identify the focus, directrix, and axis of symmetry. The

equation has the form x = 1 —

4p y2, where p = −

1 —

4 . The

focus is (p, 0), or ( − 1 —

4 , 0 ) . The directrix is x = −p, or

x = 1 —

4 . Because y is squared, the axis of symmetry is

the x-axis, y = 0.


that it is easier to substitute y-values and solve for x.

Opposite y-values result in the same x-value.

y 0 ±1 ±2 ±3 ±4

x 0 −1 −4 −9 −16

2

3

1

−3

−4

−2

21−2−1−3−4−5−6−7−8−9−10 x

y

(−14—, 0) y = 0

x = 14—

84. Step 1 Rewrite the equation in standard form.

y2 = 4x

x = 1 —

4 y2

Step 2 Identify the focus, directrix, and axis of symmetry. The

equation has the form x = 1 —

4p y2, where p = 1. The focus

is (p, 0) or (1, 0). The directrix is x = −p, or x = −1.

Because y is squared, the axis of symmetry is the x-axis,

y = 0.


that it is easier to substitute y-values and solve for x.

Opposite y-values result in the same x-value.

y 0 ±1 ±2 ±3 ±4

x 0 1 —

4 1

9 —

4 4

x

y

4

5

6

2

3

1

4 5 6 7 8 9 10321−2

y = 0

x = −1

(1, 0)

85. g(x) = −f(x) 86. g(x) = f(x) − 3

= −(x4 − 3x2 − 2x) = (x3 − x) − 3

= −x4 + 3x2 + 2x = x3 − x − 3

The graph of g is a The graph of g is a

refl ection in the x-axis translation 3 units down

of the graph of f. of the graph of f.


Chapter 5

87. g(x) = f(x − 2)

= (x − 2)3 − 4

The graph of g is a translation 2 units right of the graph of f.

88. g(x) = f(2x)

= (2x)4 + 2(2x)3 − 4(2x)2

= 16x4 + 16x3 − 16x 2

The graph of g is a horizontal shrink by a factor of 1 —

2 of the

graph of f.


1. a. D; The domain does not include negative x-values and

rises faster than graph A; The domain is x ≥ 0 and the

range is y ≥ 0.

b. C; The domain includes negative x-values and rises faster

than graph B; The domain is all real numbers and the

range is all real numbers.

c. A; The domain does not include negative x-values and

rises slower than graph D; The domain is x ≥ 0 and the

range is y ≥ 0.

d. B; The domain includes negative x-values and rises slower

than graph C; The domain is all real numbers and the

range is all real numbers.

2. a. A; The function is a translation 2 units left of the parent

square root function; The domain is x ≥ −2 and the range

is y ≥ 0.

b. D; The function is a translation 2 units right of the parent

square root function; The domain is x ≥ 2 and the range

is y ≥ 0.

c. B; The function is a translation 2 units left and 2 units down

of the parent square root function; The domain is

x ≥ −2 and the range is y ≥ −2.

d. C; The function is a refl ection in the x-axis and a

translation 2 units to the left of the parent square root

function; The domain is x ≥ −2 and the range is y ≤ 0.

3. To identify the domain and range of a radical function, use

the graph of the function.

4. When the index is odd, the domain and range of a radical

function are all real numbers. When the index is even, the

domain is based on having the expression inside of the

radical being greater than or equal to 0, and the range is

restricted as well.


1. Make a table of values and sketch the graph.

x −1 0 1 2 3

y 0 1 1.41 1.73 2

The radicand of a square root must be nonnegative. So, the

domain is x ≥ −1. The range is y ≥ 0.

x

g

y4

2

3

−2

−3

−4

4321−2−1−3−4

2. Notice that the function is of the form g(x) = − 3 √—

x + k,

where k = −2. So, the graph of g is a refl ection in the x-axis

followed by a translation 2 units down of the graph of f.

x

y

2

3

1

−3

−4

−5

−2

321−1

f

g

3. N(d) = 2.4 ⋅ E(d)

= 2.4 ⋅ 0.25 √—

d

= 0.6 √—

d

Next, fi nd N(25).

N(25) = 0.6 √—

25 = 0.6(5) = 3

It takes a dropped object 3 seconds to fall 25 feet on the Moon.

4. The transformation is not the same if you perform the

translation followed by the horizontal shrink. The rule for the

function when the order is changed is g(x) = 3 √—

6x + 3 .

5. Step 1 Solve for y.

−4y2 = x + 1

y2 = − 1 — 4 (x + 1)

y = ± √— − 1 — 4 (x + 1)

y = ± 1 —

2 √—

−(x + 1)


Chapter 5

Step 2 Graph both radical functions.

y1 = 1 —

2 √—

−(x + 1)

y2 = − 1 — 2 √—

−(x + 1)

2

−6

6

−10

y = ± −0.25x − 0.25

The vertex is (−1, 0) and the parabola opens left.


x2 + y2 = 25

y2 = 25 − x2

y = ± √—

25 − x2

Step 2 Graph both radical functions using a square viewing

window.

y1 = √—

25 − x2

y2 = − √—

25 − x2

9

−6

6

−9

y = ± 25 − x2

The radius is 5 units. The x-intercepts are ±5. The

y-intercepts are ±5.

5.3 Exercises (pp.256–258)


1. Square root functions and cube root functions are examples

of radical functions.

2. When graphing y = a 3 √—

x − h + k, translate the graph of

y = a 3 √—

x h units horizontally and k units vertically.


3. B; The function is a translation 3 units left of the parent

square root function. The domain is x ≥ −3 and the range is

y ≥ 0.

4. D;The function is a translation 3 units up of the parent square

root function. The domain is x ≥ 0 and the range is y ≥ 3.

5. F; The function is a translation 3 units right of the parent

square root function. The domain is x ≥ 3 and the range is y

≥ 0.

6. A; The function is a translation 3 units down of the parent

square root function. The domain is x ≥ 0 and the range

is y ≥ −3.

7. E; The function is a translation 3 units down and 3 units left

of the parent square root function. The domain is x ≥ −3

and the range is y ≥ −3.

8. C; The function is a translation 3 units up and 3 units right of

the parent square root function. The domain is x ≥ 3 and the

range is y ≥ 3.


x 0 1 2 3 4

y 4 5 5.41 5.73 6

x

y

4

5

6

7

8

2

3

1

−2

4 5 6 7 8321−2−1

h


domain is x ≥ 0. The range is y ≥ 4.


x 0 1 2 3 4

y −5 −4 −3.59 −3.27 −3

x

y2

1

−3

−4

−5

−6

−2

4 5 6321−2−1

g


domain is x ≥ 0. The range is y ≥ −5.


x −4 −2 0 2 4

y 2 1.59 0 −1.59 −2

x

y

2

3

−3

−2

321−2−1−3

g

The radicand of a cube root can be any real number. So, the

domain and range are all real numbers.


Chapter 5


x −4 −2 0 2 4

y 2.71 2.15 0 −2.15 −2.71

x

y

2

3

4

−3

−4

−2

4321−2−1−3−4

f




x 3 4 5 6 7

y 0 0.2 0.28 0.35 0.4

x

y

2

3

4

1

503010−1

g




x −8 −6 −4 −2 0 2 4

y −0.63 0 0.63 0.79 0.90 1 1.08

x

y

2

3

4

−2

−3

−4

−2−4−6

f




x 0 1 2 3 4

y 3 5.45 6.46 7.24 7.89

x

y

2

1

3

4

5

6

7

8

9

10

−2

4 5 6 7 8 9 10321−2−1

f




x −4 −2 0 2 4

y 4.32 3 −3 −4.32 −5.13

x

y

2

1

3

4

5

6

−3

−4

−5

−6

4 5 6321−2−3−4−5−6

g




x 0 1 2 3 4

y 0 −1 −1.19 −1.32 −1.41

x

y1

−3

−4

−2

4321−1

h

The radicand of a fourth root must be nonnegative. So, the

domain is x ≥ 0. The range is y ≤ 0.


Chapter 5


x −4 −2 0 2 4

y −1.52 −1.32 0 1.32 1.52

x

y

2

3

4

1

−2

−3

−4

4321−2−1−3−4

h

The radicand of a fi fth root can be any real number. So, the


19. Notice that the function is of the form g(x) = √—

x − h + k,

where h = −1 and k = 8. So, the graph of g is a translation

1 unit left and 8 units up of the graph of f.

x

y

2

1

3

4

5

6

7

8

10

11

12

−2

4 5 6 7 8 9 10 11 12321−2−1

g

f

20. Notice that the function is of

x

y

2

1

3

4

5

6

−2

4 5 6321−2−1

g

f

the form g(x) = a √—

x − h , where a = 2 and h = 1. So,

the graph of g is a translation

1 unit right followed by a

vertical stretch by a factor

of 2 of the graph of f.

21. Notice that the function is of

x

y

2

1

3

4

−3

−4

−2

4321−2−1−3−4

g

f

the form g(x) = − 3 √—

x + k,

where k = −1. So, the graph

of g is a refl ection in the

x-axis followed by a

translation 1 unit down

of the graph of f.

22. Notice that the function is of the form g(x) = 3 √—

x − h + k,

where h = −4 and k = −5. So, the graph of g is a

translation 4 units left and 5 units down of the graph of f.

x

y

2

1

3

4

−3

−4

−5

−6

−7

−8

−2

4321−2−1−3−4−5−6−7−8

g

f

23. Notice that the function is of the form g(x) = a(−x)1/2,

where a = 1 —

4 . So, the graph of g is a refl ection in the y-axis

followed by a vertical shrink by a factor of 1 —

4 of the graph of f.

x

y

2

1

3

4

4321−2−1−3−4

f

g

24. Notice that the function is of the form g(x) = ax1/2 + k, where

a = 1 —

3 and k = 6. So, the graph of g is a vertical shrink by a

factor of 1 —

3 followed by a translation 6 units up of the graph of f.

x

y

2

1

3

4

5

7

8

9

−3

−2

4 5 6321−2−1−3−4−5−6

f

g

25. Notice that the function is of the form g(x) = a 4 √—

x − h + k,

where a = 2, h = −5, and k = −4. So the graph of g is a

vertical stretch by a factor of 2 followed by a translation

5 units left and 4 units down of the graph of f.

x

y

2

1

3

−3

−4

−2

−2−1−3−4−5g

f


Chapter 5

26. Notice that the function is of the form g(x) = 5 √—

ax + k,

where a = −32 and k = 3. So, the graph of g is a horizontal

shrink by a factor of 1 —

32 and a refl ection in the y-axis followed

by a translation 3 units up of the graph of f.

x

y

2

1

3

4

6

7

8

−3

−4

−2

4 5 6321−2−1−3−4−5−6

g

f

27. The graph was translated

x

y2

−2

−4

6 82−2

2 units left but it should be

translated 2 units right.

28. The function is a horizontal stretch, not a horizontal shrink.

“The graph of g is a horizontal stretch by a factor of 2 and a

translation 3 units up of the parent square root function.”

29.

6

−2

−6

8 30.

6

−2

−6

8

The domain is x ≤ −1 and The domain is x ≤ 0 and

x ≥ 0. The range is y ≥ 0. x ≥ 2. The range is y ≥ 0.

31.

6

−2

−6

4

The domain is all real numbers. To fi nd the range, fi nd the

x-coordinate of the vertex of the polynomial under the radical.

x = − b —

2a = −

1 —

2(1) = −

1 —

2

Substitute into the function to fi nd the vertex.

f ( − 1 —

2 ) =

3 √——

( − 1 —

2 ) 2 + ( −

1 —

2 ) =

3

√—

( − 1 —

4 ) ( 2 —

2 ) = −

3 √—

2 —

2

So, the range of the function is y ≥ − 3 √—

2 —

2 .

32.

6

−2

−6

4



x = − b —

2a = −

(− 1) —

2(3) =

1 —

6


f ( 1 — 6 ) =

3

√—

3 ( 1 — 6 ) 2 −

1 —

6 = 3

√——

( − 1 —

12 ) ( 144

— 144

) = − 3 √—

144 —

12

So, the range of the function is y ≥ − 3 √—

144 —

12 .

33.

4

−2

−4

6



x = − b —

2a = −

1 —

2(2) = −

1 —

4


f ( − 1 —

4 ) = √

——

2 ( − 1 —

4 ) 2 + ( −

1 —

4 ) + 1 = √

—

7 —

8 ( 2 — 2 ) =

√—

14 —

4

So, the range of the function is y ≥ √

— 14 —

4 .

34.

6

−2

−6

4



x = − b —

2a = −

(−3) —

2 ( 1 — 2 ) = 3


h(3) = 3 √——

1 —

2 (3) − 3(3) + 4 =

3

√—

( − 1 —

2 ) ( 4 —

4 ) = −

3 √—

4 —

2

So, the range of the function is y ≥ − 3

√—

4 —

2 .

35. The domain of the function y = a √—

x is always x ≥ 0.

36. The range of the function y = a √—

x is sometimes y ≥ 0.

37. The domain and range of the function y = 3 √—

x − h + k are

always all real numbers.

38. The domain of the function y = a √—

(−x) + k is never x ≥ 0.


Chapter 5

39. M(n) = 0.75 ⋅ E(n)

= 0.75 ⋅ 1.22 √—

n

= 0.915 √—

n

Next, fi nd M(10,000).

M(10,000) = 0.915 √—

10,000 = 0.915(100) = 91.5

A pilot at 10,000 feet can see about 91.5 miles on the Moon.

40. s(K) = v(K)

— 1.944

= 643.855

— 1.944

√— K —

273.15

Next, fi nd s(305).

s(305) = 643.855

— 1.944

√— 305

— 273.15

≈ 350

The speed is about 350 meters per second when the air

temperature is 305 kelvin.

41. Step 1 First write a function h that represents the vertical

stretch of f.

h(x) = 2f(x)

= 2 ( √—

x + 3 )

= 2 √—

x + 6

Step 2 Then write a function g that represents the translation

of h.

g(x) = h(x) + 2

= (2 √—

x + 6) + 2 = 2 √

— x + 8

The transformed function is g(x) = 2 √—

x + 8.

42. Step 1 First write a function h that represents the refl ection of f.

h(x) = f(−x)

= 2 3 √—

(−x) − 1

= 2 3 √—

−x − 1


of h.

g(x) = h(x + 1)

= 2 3 √——

(−x + 1) − 1

= 2 3 √—

−x

The transformed function is g(x) = 2 3 √—

−x .

43. Step 1 First write a function h that represents the horizontal

shrink of f.

h(x) = f ( 3 — 2 x )

= √— 6 ( 3 — 2 x )

= √—

9x


of h.

g(x) = h(x + 4)

= √—

9(x + 4)

= √—

9x + 36

The transformed function is g(x) = √—

9x + 36 .

44. Step 1 First write a function h that represents the

translations of f.

h(x) = f ( x − 5 ) − 1

= ( − 1 — 2 4 √—

x − 5 + 3 —

2 ) − 1

= − 1 — 2 4 √—

x − 5 + 1 —

2

Step 2 Then write a function g that represents the refl ection

of h.

g(x) = −h(x)

= − ( − 1 — 2 4 √—

x − 5 + 1 —

2 )

= 1 —

2 4 √—

x − 5 − 1 —

2

The transformed function is g(x) = 1 —

2 4 √—

x − 5 − 1 —

2 .

45. Step 1 First write a function h that represents the vertical

shrink of f.

h(x) = 2f(x)

= 2 √—

x


of h.

g(x) = h(x + 1)

= 2 √—

(x + 1)

= 2 √—

x + 1


x + 1 .

46. Step 1 First write a function h that represents the refl ection

of f.

h(x) = f(−x)

= 3 √—

−x


of h.

g(x) = h(x − 2)

= 3 √—

−(x − 2)

= − 3 √—

x − 2

The transformed function is g(x) = − 3 √—

x − 2 .

47. The graph of g is a translation 3 units left of the graph of f.

g(x) = f(x + 3)

= 2 √—

x + 3

48. The graph of g is a refl ection in the x-axis followed by a

translation 9 units up of the graph of f.

g(x) = −f(x) + 9

= − 1 — 3 √—

x − 1 + 9


Chapter 5

49. The graph of g is a refl ection in the x-axis and a vertical

stretch by a factor of 2 followed by a translation 5 units left

of the graph of f.

g(x ) = −2f(x + 5)

= −2 ( − √——

(x + 5)2 − 2 ) = 2 √

—— (x + 5)2 − 2

= 2 √——

x2 + 10x + 25 − 2

= 2 √——

x2 + 10x + 23

50. The graph of g is a refl ection in the y-axis and a vertical

shrink by a factor of 4 followed by a translation 6 units up

of the graph of f.

g(x) = f(−x) + 6

= 1 —

4 3 √——

(−x)2 + 10(−x) + 6

= 1 —

4 3 √—

x2 − 10x + 6


1 — 4 y2 = x

y2 = 4x

y = ±2 √—

x

Step 2 Graph both radical

5

−5

5

−1

y = ±2 xfunctions.

y1 = 2 √—

x

y2 = −2 √—

x

The vertex is (0, 0) and the

parabola opens right.


3y2 = x

y2 = 1 —

3 x

y = ± √—

1 —

3 x


5

−2

2

−1

y = ± x3functions.

y1 = √—

1 —

3 x

y2 = − √—

1 —

3 x




−8y2 + 2 = x

−8y2 = x − 2

y2 = − 1 — 8 (x − 2)

y = ± 1 —

2 √— −

1 — 2 (x − 2)


3

−2

2

−3

y = ± 2 − x8functions.

y1 = 1 —

2 √— −

1 — 2 (x − 2)

y2 = − 1 — 2 √—

− 1 — 2 (x − 2)


parabola opens left.


2y2 = x − 4

y2 = 1 —

2 x − 2

y = ± √— 1 —

2 x − 2


11

−4

4

−1

y = ± x − 42functions.

y1 = √— 1 —

2 x − 2

y2 = − √— 1 —

2 x − 2




x + 8 = 1 —

5 y2

5x + 40 = y2

y = ± √—

5x + 40


6

−8

8

−10

y = ± 5x + 40

functions.

y1 = √—

5x + 40

y2 = − √—

5x + 40

The vertex is (−8, 0) and the



1 — 2 x = y2 − 4

1 — 2 x + 4 = y2

y = ± √— 1 —

2 x + 4


2

−4

4

−10

y = ± x + 412functions.

y1 = √— 1 —

2 x + 4

y2 = − √— 1 —

2 x + 4




Chapter 5


x2 + y2 = 9

y2 = 9 − x2

y = ± √—

9 − x2


6

−4

4

−6

y = ± 9 − x2

functions using a square

viewing window.

y1 = √—

9 − x2

y2 = − √—

9 − x2

The radius is 3 units.

The x-intercepts are ±3. The y-intercepts are ±3.


x2 + y2 = 4

y2 = 4 − x2

y = ± √—

4 − x2


6

−4

4

−6

y = ± 4 − x2


viewing window.

y1 = √—

4 − x2

y2 = − √—

4 − x2




1 − y2 = x2

y2 + x2 = 1

y = ± √—

1 − x2


6

−4

4

−6

y = ± 1 − x2


viewing window.

y1 = √—

1 − x2

y2 = − √—

1 − x2

The radius is 1 unit.



64 − x2 = y2

y = ± √—

64 − x2


15

−10

10

−15

y = ± 64 − x2

functions using a

square viewing

window.

y1 = √—

64 − x2

y2 = − √—

64 − x2




−y2 = x2 − 36

y2 = 36 − x2

y = ± √—

36 − x2


12

−8

8

−12

y = ± 36 − x2

functions using a

square viewing

window.

y1 = √—

36 − x2

y2 = − √—

36 − x2




x2 = 100 − y2

y2 + x2 = 100

y = ± √—

100 − x2


18

−12

12

−18

y = ± 100 − x2


viewing window.

y1 = √—

100 − x2

y2 = − √—

100 − x2



63.

ℓ

T

2

1

3

4

5

4 5321

The length of a pendulum that has a period of 2 seconds is

about 3 feet. Sample answer: Locate the T-value 2 on the

graph and estimate theℓ-value.

64. The graph represents a square root function because there is

no left half. The domain of the function is x ≥ −3. The range

of the function is y ≥ 1.

65.

p

s

100

50

150

200

250

800400

a. When the speed of a 3500-pound car is 200 miles per

hour, the power is about 2468 horsepower.

b. The average rate of change is

s(1500) − s(1000)

—— 1500 − 1000

≈ 169.4 − 148

—— 500

≈ 0.04.

So, the average rate of change in the speed from

1000 horsepower to 1500 horsepower is about

0.04 mile per hour per horsepower.


Chapter 5

66. Sample answer: f can be written as f(x) = √—

−x + 4 or

f(x) = − √—

x − 3 + 1.

g(x) = √——

−(x + 2) + 4 + 1

= √——

−x − 2 + 4 + 1 = √—

−x + 2 + 1

h(x) = − √—

x + 2 − 3 + 1 + 1 = − √—

x − 1 + 2

67. a. Create a table of values for the 140-pound skydiver.

Compare the values to the table for the 165-pound

skydiver.

A 1 3 5 7

Vt 398.7 230.2 178.3 150.7

The 165-pound skydiver has a greater terminal velocity

for each value of A.

b. When A = 1, the diver is most likely vertical. When

A = 7, the diver is most likely horizontal.

68. a. S = πr + πr2

πr + πr2 = S

r + r2 = 1 —

π S

r2 + r + ( 1 — 2 )

2

= 1 —

π S + ( 1 —

2 )

2

( r + 1 —

2 ) 2 =

1 —

π S +

1 —

4

r + 1 —

2 =

1 —

√—

π √— S +

π — 4

r = 1 —

√—

π √— S +

π — 4 −

1 —

2

Notice only the positive form is used because r represents

a length, which cannot be negative.

b.

20

1

0 π

12y = x + − 1

4

When the surface area is 3π — 4 square units, the radius is

0.5 unit.


69. ∣ 3x + 2 ∣ = 5

3x + 2 = 5 or 3x + 2 = −5

3x = 3 or 3x = −7

x = 1 or x = − 7 — 3

So, the solutions are x = 1 and x = − 7 — 3 .

70. The equation has no solution because the left-hand side is

positive and the right-hand side is negative.

71. ∣ 2x − 6 ∣ = ∣ x ∣ 2x − 6 = x or 2x − 6 = −x

x = 6 or 3x = 6

x = 6 or x = 2

So, the solutions are x = 6 and x = 2.

72. ∣ x + 8 ∣ = ∣ 2x + 2 ∣ x + 8 = 2x + 2 or x + 8 = −(2x + 2)

−x = −6 or x + 8 = −2x − 2

x = 6 or 3x = −10

x = 6 or x = − 10

— 3

So, the solutions are x = 6 and x = − 10

— 3 .

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

x2 + 7x + 12 = 0

(x + 3)(x + 4) = 0

x = −3 or x = −4

The numbers −3 and −4 are critical values of the original

inequality. Plot −3 and −4 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 determine whether it

satisfi es the inequality.

Test x = −5. Test x = 0.

Test x = − .

−5 −4 −3 −2 −1 0−672

(−5)2 + 7(−5) + 12 < 0

( − 7 — 2 )

2 + 7 ( −

7 — 2 ) + 12 < 0

(0)2 + 7(0) + 12 < 0 So, the solution is −4 < x < −3.

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

x 2 − 10x + 25 = 4

x2 − 10x + 21 = 0

(x − 3)(x − 7) = 0

x = 3 or x = 7

The numbers 3 and 7 are critical values of the original

inequality. Plot 3 and 7 on a number line, using closed dots

because the values do satisfy the inequality. The critical x-values

partition the number line into three intervals. Test an x-value in

each interval to determine whether it satisfi es the inequality.

Test x = 4.Test x = 0.

109876543210

Test x = 8.

02 − 10(0) + 25 ≥ 4

42 − 10(4) + 25 ≥ 4

82 − 10(8) + 25 ≥ 4 So, the solution is x ≤ 3 and x ≥ 7.


Chapter 5

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

2x2 + 6 = 13x

2x2 − 13x + 6 = 0

(x − 6)(2x − 1) = 0

x = 6 or x = 1 —

2

The numbers 6 and 1 —

2 are critical values of the original

inequality. Plot 1 —

2 and 6 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 determine whether it satisfi es the

inequality.

Test x = 1.

Test x = 0. 12

876543210−1−2

Test x = 7.

2(0)2 + 6 > 13(0)

2(1)2 + 6 ≥ 13(1)

2(7)2 + 6 > 13(7)

So, the solution is x < 1 —

2 and x > 6.

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

1 — 8 x2 + x = −2

1 — 8 x2 + x + 2 = 0

x2 + 8x + 16 = 0

(x + 4)2 = 0

x + 4 = 0

x = −4

The number −4 is a critical value of the original inequality.

Plot −4 on a number line, using a closed dot because

the value does satisfy the inequality. The critical x-value

partitions the number line into two intervals. Test an

x-value in each interval to determine whether it satisfi es the

inequality.

Test x = 0.

10−1−2−3−4−5−6

Test x = −5.

1 — 8 (0)2 + 0 < −2

1 — 8 (−5)2 + (−5) < −2

So, the solution is x = −4.

5.1–5.3 What Did You Learn? (p. 259)

1. Use the defi nition of a rational exponent to rewrite each

expression on the left.

2. Rewrite 2.512m1

— 2.512m2

as 2.512m1 − m2 to simplify the expression.

3. Find the distance a pilot can see to the horizon on Earth from

the same altitude as on Mars and compare the results.

4. πr2 is the area of the circular base and πr is the lateral

surface area.

5.1–5.3 Quiz (p. 260)

1. Because n = 4 is even and a = 81 > 0, 81 has two real fourth

roots. Because 34 = 81 and (−3)4 = 81, you can write

± 4 √—

81 = ±3 or ±811/4 = ±3.


real fi fth root. Because (−4)5 = −1024, you can write

5 √—

−1024 = −4 or (−1024)1/4 = −4.

3. a. 163/4 = (161/4)3 = 23 = 8

The fourth root of 16 is 2 and 23 is 8.

163/4 = (161/4)3 = 23 = 8

b. 1252/3 = (1251/3)2 = 52 = 25


4. 2x6 = 1458 5. (x + 6)3 = 28

x6 = 729 x + 6 ≈ 3.04

x = ±3 x ≈ −2.96


x = −3 and x = 3.

6. ( 481/4 —

61/4 )

6

= [ ( 48 —

6 )

1/4

] 6

7. 4 √—

3 ⋅ 4 √—

432 = 4 √—

3 ⋅ 432

= [ (8)1/4 ] 6 = 4 √—

1296

= 86/4 = 6

= 83/2

8. 1 —

3 + √—

2 =

1 —

3 + √—

2 ⋅

3 − √—

2 —

3 − √—

2

= 3 − √

— 2 ——

( 3 + √—

2 ) ( 3 − √—

2 )

= 3 − √

— 2 —

32 − ( √—

2 ) 2

= 3 − √

— 2 —

9 − 2

= 3 − √

— 2 —

7

9. 3 √—

16 − 5 3 √—

2 = 3 √—

8 ⋅ 2 − 5 3 √—

2

= 3 √—

8 ⋅ 3 √—

2 − 5 3 √—

2

= 2 3 √—

2 − 5 3 √—

2

= −3 3 √—

2

10. 8 √—

x9y8z16 = 8 √—

x9 8 √—

y8 8 √—

z16

= 8 √—

xx8 8 √—

y8 8 √—

z16

= x ∣ y ∣ z2 8 √—

x


Chapter 5

11. 3 √—

216p9 = 3 √—

216 ⋅ 3 √—

p9

= 6p3

12. 5 √—

32 —

5 √—

m3

= 2 —

(m3)1/5

= 2 —

m3/5

= 2 —

m3/5 ⋅

m2/5

— m2/5

= 2m2/5

— m

= 2

5 √—

m2 —

m

13. 4 √—

n4q + 7n 4 √—

q = n 4 √—

q + 7n 4 √—

q

= 8n 4 √—

q


x −4 −2 0 2 4

y −2.17 −1.52 1 3.52 4.17

x

y

2

1

3

4

5

6

−3

−4

−2

4 5321−2−1−3−4−5

f



15. The graph of g is a translation 1 unit left of the parent square

root function.

g(x) = f (x + 1)

= √—

x + 1

16. The graph of g is a refl ection in the x-axis followed by a

translation 1 unit up of the parent cube root function.

g(x) = −f (x) + 1

= − 3 √—

x + 1

17. The graph of g is a translation 1 unit right and 2 units down

of the parent square root function.

g(x) = f (x − 1) − 2

= √—

x − 1 − 2


x = 3y2 − 6

x + 6 = 3y2

1 — 3 x + 2 = y2

y = ± √— 1 —

3 x + 2


4

−4

4

−8

y = ± x + 63functions.

y1 = √— 1 —

3 x + 2

y2 = − √— 1 —

3 x + 2



19. Substitute 161 for V in the equation and solve for s.

161 = 0.47s3

342.6 ≈ s3

s ≈ 7

When the volume of the octahedron is 161 cubic millimeters,

the side length is about 7 millimeters.

20.

d

s

20

25

30

35

40

45

10

15

5

30 50 70 9010

Find the speed of the car

when d = 90.

s = 4 √—

90 ≈ 38

So, when the skid mark is

90 feet, the speed of the

car was about 38 miles per

hour. Therefore, the car

was going faster than the

posted speed limit of

35 miles per hour.


1. a. C; The domain of the equation is x ≥ 1. The graph crosses

the x-axis at x = 2.

Check: √—

2 − 1 − 1 ==?

0

√—

1 − 1 ==?

0

1 − 1 ==?

0

0 = 0 ✓

So, the solution is x = 2.

b. E; The domain of the equation is x ≥ −1. The graph

crosses the x-axis at x = 2.

Check: √—

2(2) + 2 − √—

2 + 4 ==?

0

√—

6 − √—

6 ==?

0

0 = 0 ✓


c. B; The domain of the equation is −3 ≤ x ≤ 3. The graph

touches the x-axis at x = −3 and x = 3.

Check: √—

9 − (−3)2 ==?

0 √—

9 − 32 ==?

0

√—

9 − 9 ==?

0 √—

9 − 9 ==?

0

√—

0 ==?

0 √—

0 ==?

0

0 = 0 ✓ 0 = 0 ✓

So, the solutions are x = −3 and x = 3.


Chapter 5

d. D; The domain of the equation is x ≥ −2. The graph

crosses the x-axis at x = 2.

Check: √—

2 + 2 − 2 ==?

0

√—

4 − 2 ==?

0

2 − 2 ==?

0

0 = 0 ✓


e. F; The domain of the equation is x ≤ 2. The graph crosses

the x-axis at x = 1.

Check: √—

−1 + 2 − 1 ==?

0

√—

1 − 1 ==?

0

1 − 1 ==?

0

0 = 0 ✓


f. A; The domain of the equation is all real numbers. The

graph does not cross the x-axis, so the equation has no

solutions.

2. a. Sample answer: Use the equation in Exploration 1(c).

x −3 −2 −1 0 1 2 3

√—

9 − x2 0 √—

5 √—

8 3 √—

8 √—

5 0

The solutions are x = 3 and x = −3.

b. Sample answer: √—

9 − x2 = 0

9 − x2 = 0

x = ±3

Isolate the radical; square both sides.

3. To solve a radical equation, isolate the radical on one side.

Use the index to determine to what exponent each side of the

equation is to be raised to eliminate the radical.

4. Sample answer: For the given equation, an analytical

approach is preferred because it gives the most accurate

solution.

√—

x + 3 − √—

x − 2 = 1

√—

x + 3 = 1 + √—

x − 2

( √—

x + 3 ) 2 = ( 1 + √—

x − 2 ) 2

x + 3 = 1 + 2 √—

x − 2 + ( √—

x − 2 ) 2

x + 3 = 1 + 2 √—

x − 2 + x − 2

x + 3 = 2 √—

x − 2 + x − 1

3 = 2 √—

x − 2 − 1

4 = 2 √—

x − 2

2 = √—

x − 2

22 = ( √—

x − 2 ) 2

4 = x − 2

6 = x



1. 3 √—

x − 9 = −6 Check: 3 √—

27 − 9 ==?

−6

3 √—

x = 3 3 − 9 ==?

−6

( 3 √—

x ) 3 = 33 −6 = −6 ✓

x = 27 The solution is x = 27.

2. √—

x + 25 = 2 Check: √—

−21 + 25 ==?

2

( √—

x + 25 ) 2 = 22 √—

4 ==?

2

x + 25 = 4 2 = 2 ✓

x = −21 The solution is x = −21.

3. 2 3 √—

x − 3 = 4 Check: 2 3 √—

11 − 3 ==?

4

3 √—

x − 3 = 2 2 3 √—

8 ==?

4

( 3 √—

x − 3 ) 3 = 23 2 ⋅ 2 ==?

4

x − 3 = 8 4 = 4


4. v(p) = 6.3 √—

1013 − p

48.3 = 6.3 √—

1013 − p

7.67 ≈ √—

1013 − p

7.672 ≈ ( √—

1013 − p ) 2

58.78 ≈ 1013 − p

−954.22 ≈ −p

954.22 ≈ p

The air pressure at the center of the hurricane is about

954 millibars.

5. √—

10x + 9 = x + 3

( √—

10x + 9 ) 2 = (x + 3)2

10x + 9 = x2 + 6x + 9

0 = x2 − 4x

0 = x(x − 4)

x = 0 or x − 4 = 0

x = 0 or x = 4

Check: √—

10(0) + 9 ==?

0 + 3

√—

9 ==?

3

3 = 3 ✓

√—

10(4) + 9 ==?

4 + 3

√—

49 ==?

7

7 = 7 ✓

The solutions are x = 0 and x = 4.

6. √—

2x + 5 = √—

x + 7 Check: √—

2(2) + 5 =?

√—

2 + 7

( √—

2x + 5 ) 2 = ( √—

x + 7 ) 2 √—

9 =?

√—

9

2x + 5 = x + 7 3 = 3 ✓



Chapter 5

7. √—

x + 6 − 2 = √—

x − 2

√—

x + 6 = √—

x − 2 + 2

( √—

x + 6 ) 2 = ( √—

x − 2 + 2 ) 2

x + 6 = ( √—

x − 2 ) 2 + 4 √—

x − 2 + 4

x + 6 = x − 2 + 4 √—

x − 2 + 4

6 = 4 √—

x − 2 + 2

4 = 4 √—

x − 2

1 = √—

x − 2

12 = ( √—

x − 2 ) 2

1 = x − 2

3 = x

Check: √—

3 + 6 − 2 ==?

√—

3 − 2

√—

9 − 2 ==?

√—

1

3 − 2 ==?

1

1 = 1 ✓


8. (3x)1/3 = −3 Check: [ 3(−9) ] 1/3 ==?

−3

[ (3x)1/3 ] 3 = (−3)3 −271/3 ==?

−3

3x = −27 −3 = −3 ✓

x = −9 The solution is x = −9.

9. (x + 6)1/2 = x

[ (x + 6)1/2 ] 2 = x2

x + 6 = x2

0 = x2 − x − 6

0 = (x − 3)(x + 2)

x − 3 = 0 or x + 2 = 0

x = 3 or x = −2

Check: (3 + 6)1/2 ==?

3 (−2 + 6)1/2 ==?

−2

91/2 ==?

3 41/2 ==?

−2

3 = 3 ✓ 2 ≠ −2 ✗


10. (x + 2)3/4 = 8 Check: (14 + 2)3/4 ==?

8

[ (x + 2)3/4 ] 4/3 = 84/3 163/4 ==?

8

x + 2 = 16 8 = 8 ✓


11. a. Solve for x.

2 √—

x − 3 ≥ 3

2 √—

x ≥ 6

√—

x ≥ 3

x ≥ 9

Consider the radicand: x ≥ 0

So, the solution is x ≥ 9.

b. Solve for x.

4 3 √—

x + 1 < 8

3 √—

x + 1 < 2

x + 1 < 8

x < 7

So, the solution is x < 7.



1. The equation is not a radical equation because the radicand

does not contain a variable.

2. First, subtract 10 from both sides of the inequality. Then

square each side. Eliminate any solutions that would make

the radicand negative.

Monitoring Progress and Modeling With Mathematics

3. √—

5x + 1 = 6 Check: √—

5(7) + 1 ==?

6

( √—

5x + 1 ) 2 = 62 √—

36 ==?

6

5x + 1 = 36 6 = 6 ✓

5x = 35


4. √—

3x + 10 = 8 Check: √—

3(18) + 10 ==?

8

( √—

3x + 10 ) 2 = 82 √—

64 ==?

8

3x + 10 = 64 8 = 8 ✓

3x = 54


5. 3 √—

x − 16 = 2 Check: 3 √—

24 − 16 ==?

2

( 3 √—

x − 16 ) 3 = 23 3 √—

8 ==?

2

x − 16 = 8 2 = 2 ✓


6. 3 √—

x − 10 = −7 Check: 3 √—

27 − 10 ==?

−7

3 √—

x = 3 3 − 10 ==?

−7

( 3 √—

x ) 3 = 33 −7 = −7 ✓


7. −2 √—

24x + 13 = −11 Check:

−2 √—

24x = −24 −2 √—

24(6) + 13 ==?

−11

√—

24x = 12 −2 √—

144 + 13 ==?

−11

( √—

24x ) 2 = 122 −2(12) + 13 ==?

−11

24x = 144 −24 + 13 ==?

−11

x = 6 −11 = −11 ✓



Chapter 5

8. 8 3 √—

10x − 15 = 17 Check: 8 3 √— 10 ( 64

— 10

) − 15 ==?

17

8 3 √—

10x = 32 8 3 √—

64 − 15 ==?

17

3 √—

10x = 4 8 ⋅ 4 − 15 ==?

17

( 3 √—

10x ) 3 = 43 32 − 15 ==?

17

10x = 64 17 = 17 ✓

x = 6.4 The solution is x = 6.4.

9. 1 — 5 3 √—

3x + 10 = 8 Check: 1 — 5 3

√— 3 ( − 1000

— 3 ) + 10 ==?

8

1 — 5 3 √—

3x = −2 1 —

5 3 √—

−1000 + 10 ==?

8

3 √—

3x = −10 1 —

5 (−10) + 10 ==

? 8

( 3 √—

3x ) 3 = (−10)3 −2 + 10 ==?

8

3x = −1000 8 = 8 ✓

x = − 1000

— 3 The solution is − 1000

— 3 .

10. √—

2x − 2 —

3 = 0 Check: √— 2 ( 2 —

9 ) −

2 —

3 ==?

0

√—

2x = 2 —

3 √—

4 —

9 −

2 —

3 ==?

0

( √—

2x ) 2 = ( 2 — 3 ) 2

2 —

3 −

2 —

3 ==?

0

2x = 4 —

9 0 = 0 ✓

x = 2 —

9 The solution is x =

2 —

9 .

11. 2 5 √—

x + 7 = 15 Check: 2 5 √—

1024 + 7 ==?

15

2 5 √—

x = 8 2 ⋅ 4 + 7 ==?

15

5 √—

x = 4 15 = 15 ✓

( 5 √—

x ) 5 = 45


12. 4 √—

4x − 13 = −15

4 √—

4 x = −2

The equation has no real solution.

13. h = 62.5 3 √

— t + 75.8

250 = 62.5 3 √

— t + 75.8

174.2 = 62.5 3 √

— t

2.79 ≈ 3 √

— t

(2.79)3 ≈ 3 √

— t 3

21.7 ≈ t

The age of a male Asian elephant that is 250 centimeters tall

is about 21.7 years.

14. v = √—

2gh

v ≈ √—

2 ⋅ 9.8h

v ≈ √—

19.6h

15 ≈ √—

19.6h

(15)2 ≈ ( √—

19.6h ) 2

225 ≈ 19.6h

11.5 ≈ h

The height at the top of the swing ride is about 11.5 meters.

15. x − 6 = √—

3x

(x − 6)2 = ( √—

3x ) 2

x2 − 12x + 36 = 3x

x2 − 15x + 36 = 0

(x − 12)(x − 3) = 0

x − 12 = 0 or x − 3 = 0

x = 12 or x = 3

Check:

12 − 6 ==?

√—

3(12) 3 − 6 ==?

√—

3(3)

6 ==?

√—

36 −3 ==?

√—

9

6 = 6 ✓ −3 ≠ 3 ✗


16. x − 10 = √—

9x

(x − 10)2 = ( √—

9x ) 2

x2 − 20x + 100 = 9x

x2 − 29x + 100 = 0

(x − 4)(x − 25) = 0

x − 4 = 0 or x − 25 = 0

x = 4 or x = 25

Check:

4 − 10 ==?

√—

9(4) 25 − 10 ==?

√—

9(25)

−6 ==?

√—

36 15 ==?

√—

225

−6 ≠ 6 ✗ 15 = 15 ✓


17. √—

44 − 2x = x − 10

( √—

44 − 2x ) 2 = (x − 10)2

44 − 2x = x2 − 20x + 100

0 = x2 − 18x + 56

0 = (x − 4)(x − 14)

x − 4 = 0 or x − 14 = 0

x = 4 or x = 14

Check:

√—

44 − 2(4) ==?

4 − 10

√—

36 ==?

−6

6 ≠ −6 ✗

√—

44 − 2(14) ==?

14 − 10

√—

16 ==?

4

4 = 4 ✓



Chapter 5

18. √—

2x + 30 = x + 3

( √—

2x + 30 ) 2 = (x + 3)2

2x + 30 = x2 + 6x + 9

0 = x2 + 4x − 21

0 = (x + 7)(x − 3)

x + 7 = 0 or x − 3 = 0

x = −7 or x = 3

Check:

√—

2(−7) + 30 ==?

−7 + 3 √—

2(3) + 30 ==?

3 + 3

√—

16 ==?

−4 √—

36 ==?

6

4 ≠ −4 ✗ 6 = 6 ✓


19. 3 √—

8x3 − 1 = 2x − 1

( 3 √—

8x3 − 1 ) 3= (2x − 1)3

8x3 − 1 = 8x3 − 12x2 + 6x − 1

0 = −12x2 + 6x

0 = −6x(2x − 1)

−6x = 0 or 2x − 1 = 0

x = 0 or x = 1 —

2

Check:

3 √—

8(0)3 − 1 ==?

2(0) − 1 3 √— 8 ( 1 —

2 ) 3 − 1 ==

? 2 ( 1 —

2 ) − 1

3 √—

−1 ==?

−1 √—

1 − 1 ==?

1 − 1

−1 = −1 ✓ 0 = 0 ✓

The solutions are x = 0 and x = 1 —

2 .

20. 4 √—

3 − 8x2 = 2x

( 4 √—

3 − 8x2 ) 4 = (2x)4

3 − 8x2 = 16x4

0 = 16x4 + 8x2 − 3

0 = (4x2 − 1)(4x2 + 3)

0 = (2x + 1)(2x − 1)(4x2 + 3)

2x + 1 = 0 or 2x − 1 = 0 or 4x2 + 3 = 0

x = − 1 — 2 or x =

1 —

2 or No real solution

Check:

4 √— 3 − 8 ( −

1 — 2 )

2 ==?

2 ( − 1 — 2 ) 4

√— 3 − 8 ( 1 — 2 ) 2 ==? 2 ( 1 —

2 )

4 √—

1 ==?

−1 4 √—

1 ==?

1

1 ≠ −1 ✗ 1 = 1 ✓

The solution is x = 1 —

2 .

21. √—

4x + 1 = √—

x + 10 Check:

( √—

4x + 1 ) 2 = ( √

— x + 10 ) 2

√—

4(3) + 1 ==?

√—

3 + 10

4x + 1 = x + 10 √—

13 = √—

13 ✓

3x = 9


22. √—

3x − 3 − √—

x + 12 = 0

√—

3x − 3 = √—

x + 12

( √—

3x − 3 ) 2 = ( √—

x + 12 ) 2

3x − 3 = x + 12

2x = 15

x = 15

— 2

Check: √— 3 ( 15 —

2 ) − 3 − √—

15 —

2 + 12 ==

? 0

√—

39

— 2 − √—

39

— 2 ==

? 0

0 = 0 ✓

The solution is x = 15

— 2 .

23. 3 √—

2x − 5 − 3 √—

8x + 1 = 0

3 √—

2x − 5 = 3 √—

8x + 1

( 3 √—

2x − 5 ) 3 = ( 3 √—

8x + 1 ) 3

2x − 5 = 8x + 1

−6x = 6

x = −1

Check: 3 √—

2(−1) − 5 − 3 √—

8(−1) + 1 ==?

0

3 √—

−7 − 3 √—

−7 ==?

0

0 = 0 ✓

The solution is x = −1.

24. 3 √—

x + 5 = 2 3 √—

2x + 6

( 3 √—

x + 5 ) 3 = ( 2 3 √—

2x + 6 ) 3

x + 5 = 8(2x + 6)

x + 5 = 16x + 48

−15x = 43

x = − 43

— 15

Check: 3 √— −

43 — 15 + 5 ==

? 2 3

√— 2 ( − 43

— 15 ) + 6

3 √—

32

— 15

==?

2 3 √—

4 —

15

2 3 √—

4 —

15 = 2 3

√—

4 —

15 ✓

The solution is x = − 43

— 15 .


Chapter 5

25. √—

3x − 8 + 1 = √—

x + 5

( √—

3x − 8 + 1 ) 2 = ( √—

x + 5 ) 2

( √—

3x − 8 ) 2 + 2 √—

3x − 8 + 1 = x + 5

3x − 8 + 2 √—

3x − 8 + 1 = x + 5

3x − 7 + 2 √—

3x − 8 = x + 5

2 √—

3x − 8 = −2x + 12

√—

3x − 8 = −x + 6

( √—

3x − 8 ) 2 = (−x + 6)2

3x − 8 = x2 − 12x + 36

0 = x2 − 15x + 44

0 = (x − 11)(x − 4)

x − 11 = 0 or x − 4 = 0

x = 11 or x = 4

Check: √—

3(11) − 8 + 1 ==?

√—

11 + 5

√—

25 + 1 ==?

√—

16

5 + 1 ==?

4

6 ≠ 4 ✗

√—

3(4) − 8 + 1 ==?

√—

4 + 5

√—

4 + 1 ==?

√—

9

2 + 1 ==?

3

3 = 3 ✓


26. √—

x + 2 = 2 − √—

x

( √—

x + 2 ) 2 = ( 2 − √

— x ) 2

x + 2 = 4 − 4 √—

x + ( √—

x ) 2

x + 2 = 4 − 4 √—

x + x

2 = 4 − 4 √—

x

−2 = −4 √—

x

1 — 2 = √

— x

( 1 — 2 ) 2 = ( √

— x ) 2

1 — 4 = x

Check: √— 1 —

4 + 2 ==

? 2 − √—

1 —

4

√—

9 — 4 ==

? 2 − 1 —

2

3 — 2 = 3 —

2 ✓

The solution is x = 1 —

4 .

27. 2x2/3 = 8 Check:

x2/3 = 4 2(8)2/3 ==?

8 2(−8)2/3 ==?

8

(x2/3)3/2 = 43/2 2 ⋅ 4 ==?

8 2 ⋅ 4 ==?

8

x = ±8 8 = 8 ✓ 8 = 8 ✓

The solutions are x = ±8.

28. 4x3/2 = 32 Check: 4(4)3/2 ==?

32

x3/2 = 8 4 ⋅ 8 ==?

32

(x3/2)2/3 = 82/3 32 = 32 ✓


29. x1/4 + 3 = 0 Check: 811/4 + 3 ==?

0

x1/4 = −3 3 + 3 ==?

0

(x1/4)4 = (−3)4 6 ≠ 0 ✗

x = 81 The equation has no real solution.

30. 2x3/4 − 14 = 40 Check:

2x3/4 = 54 2(81)3/4 − 14 ==?

40

x3/4 = 27 2(27) − 14 ==?

40

(x3/4)4/3 = 274/3 40 = 40 ✓


31. (x + 6)1/2 = x

[ (x + 6)1/2 ] 2 = x2

x + 6 = x2

0 = x2 − x − 6

0 = (x − 3)(x + 2)

x − 3 = 0 or x + 2 = 0

x = 3 or x = −2

Check:

(3 + 6)1/2 ==?

3 (−2 + 6)1/2 ==?

−2

91/2 ==?

3 41/2 ==?

−2

3 = 3 ✓ 2 ≠ −2 ✗


32. (5 − x)1/2 − 2x = 0

(5 − x)1/2 = 2x

[ (5 − x)1/2 ] 2 = (2x)2

5 − x = 4x2

0 = 4x2 + x − 5

0 = (x − 1)(4x + 5)

x − 1 = 0 or 4x + 5 = 0

x = 1 or x = − 5 — 4

Check:

(5 − 1)1/2 − 2(1) ==?

0 ( 5 − ( − 5 — 4 ) ) 1/2

− 2 ( − 5 — 4 ) ==

? 0

41/2 − 2 ==?

0 ( 25 —

4 ) 1/2

+ 5 —

2 ==?

0

2 − 2 ==?

0 5 —

2 +

5 —

2 ==?

0

0 = 0 ✓ 5 ≠ 0 ✗


262 Algebra 2 Copyright © Big Ideas Learning, LLC. Worked-Out Solutions All rights reserved.

Chapter 5

33. 2(x + 11)1/2 = x + 3

[ 2(x + 11)1/2 ] 2 = (x + 3)2

4(x + 11) = x2 + 6x + 9

4x + 44 = x2 + 6x + 9

0 = x2 + 2x − 35

0 = (x + 7)(x − 5)

x + 7 = 0 or x − 5 = 0

x = −7 or x = 5

Check:

2(−7 + 11)1/2 ==?

−7 + 3 2(5 + 11)1/2 ==?

5 + 3

2(4)1/2 ==?

−4 2(16)1/2 ==?

8

2 ⋅ 2 ==?

−4 2(4) ==?

8

4 ≠ −4 ✗ 8 = 8 ✓


34. (5x2 − 4)1/4 = x

[ (5x2 − 4)1/4 ] 4 = x4

5x2 − 4 = x4

0 = x4 − 5x2 + 4

0 = (x2 − 1)(x2 − 4)

0 = (x + 1)(x − 1)(x + 2)(x − 2)

x + 1 = 0 or x − 1 = 0

x = −1 or x = 1

or

x + 2 = 0 or x − 2 = 0

x = −2 or x = 2

Check:

[ 5(−1)2 − 4 ] 1/4 ==?

−1 [ 5(1)2 − 4 ] 1/4 ==?

1

11/4 ==?

−1 11/4 ==?

1

1 ≠ −1 ✗ 1 = 1 ✓

[ 5(−2)2 − 4 ] 1/4 ==?

−2 [ 5(2)2 − 4 ] 1/4 ==?

2

161/4 ==?

−2 161/4 ==?

2

2 ≠ −2 ✗ 2 = 2 ✓


35. The right-hand side was 36. When raising each side to

not cubed. an exponent, the 8 was

not included.

3 √—

3x − 8 = 4 8x3/2 = 1000

( 3 √—

3x − 8 ) 3 = 43 (8x3/2)2/3 = 10002/3

3x − 8 = 64 4x = 100

3x = 72 x = 25

x = 24

37. 2 3 √—

x − 5 ≥ 3 38. 3 √—

x − 4 ≤ 5

2 3 √—

x ≥ 8 ( 3 √—

x − 4 ) 3 ≤ 53

3 √—

x ≥ 4 x − 4 ≤ 125

( 3 √—

x ) 3 ≥ 43 x ≤ 129

x ≥ 64 The solution is x ≤ 129.

The solution is x ≥ 64.

39. Solve for x. 40. Solve for x.

4 √—

x − 2 > 20 7 √—

x + 1 < 9

√—

x − 2 > 5 7 √—

x < 8

x − 2 > 25 √—

x < 8 —

7

x > 27 x < 64

— 49

Consider the radicand. Consider the radicand.

x − 2 ≥ 0 x ≥ 0

x ≥ 2 So, the solution is

So, the solution is x > 27. 0 ≤ x < 64

— 49

.

41. Solve for x. 42. 3 √—

x + 7 ≥ 3

2 √—

x + 3 ≤ 8 ( 3 √—

x + 7 ) 3 ≥ 33

2 √—

x ≤ 5 x + 7 ≥ 27

√—

x ≤ 5 —

2 x ≥ 20

x ≤ 25

— 4 The solution is x ≥ 20.

Consider the radicand.

x ≥ 0

So, the solution is 0 ≤ x ≤ 25

— 4 .

43. −2 3 √—

x + 4 < 12 44. Solve for x.

3 √—

x + 4 > −6 −0.25 √—

x − 6 ≤ −3

( 3 √—

x + 4 ) 3 > (−6)3 −0.25 √

— x ≤ 3

x + 4 > −216 √—

x ≥ −12

x > −220 Consider the radicand.

The solution is x > −220. x ≥ 0


45. ℓ= 54d3/2

3 = 54d3/2

0.056 ≈ d3/2

(0.056)2/3 ≈ (d3/2)2/3

0.15 ≈ d

The diameter of a standard nail that is 3 inches long is about

0.15 inch.

46. a. Snowboarder: Kangaroo:

1.21 = 0.5 √—

h 0.81 = 0.5 √—

h

2.42 = √—

h 1.62 = √—

h

5.9 ≈ h 2.6 ≈ h

The height of the The height of the kangaroo

snowboarder is about is about 2.6 feet.

5.9 feet.

Copyright © Big Ideas Learning, LLC. Algebra 2 263All rights reserved. Worked-Out Solutions

Chapter 5

b. Snowboarder: Kangaroo:

2.42 = 0.5 √—

h 1.62 = 0.5 √—

h

4.84 = √—

h 3.24 = √—

h

23.4 ≈ h 10.5 ≈ h

The height of the The height of the kangaroo

snowboarder is about is about 10.5 feet.

23.4 feet.

c. no; When the time is doubled, the height increases by a

factor of 4.

47. Substitute y for x − 3 in Equation 1 and solve for y.

y2 = y

y2 − y = 0

y(y − 1) = 0

y = 0 or y − 1 = 0

y = 0 or y = 1

Substitute the values for y into Equation 2 and solve for x.

y = 0: 0 = x − 3 y = 1: 1 = x − 3

3 = x 4 = x

The solutions are (3, 0) and (4, 1).

x

y4

2

4 6

(4, 1)(3, 0)

48. Substitute x + 5 for y in Equation 1 and solve for x.

(x + 5)2 = 4x + 17

x2 + 10x + 25 = 4x + 17

x2 + 6x + 8 = 0

(x + 2)(x + 4) = 0

x + 2 = 0 or x + 4 = 0

x = −2 or x = −4

Substitute the values for x into Equation 2 and solve for y.

x = −2: y = −2 + 5 x = −4: y = −4 + 5

y = 3 y = 1

The solutions are (−2, 3) and (−4, 1).

x

y

2

−2−4

(−2, 3)

(−4, 1)

49. Substitute x − 2 for y in Equation 1 and solve for x.

x 2 + (x − 2)2 = 4

x 2 + x 2 − 4x + 4 = 4

2x 2 − 4x + 4 = 4

2x 2 − 4x = 0

2x(x − 2) = 0

2x = 0 or x − 2 = 0

x = 0 or x = 2

Substitute the values for x into Equation 2 and solve for y.

x = 0: y = 0 − 2 x = 2: y = 2 − 2

y = −2 y = 0

The solutions are (0, −2) and (2, 0).

x

y

1

3

31−1−3

(0, −2)

(2, 0)

50. Substitute − 3 — 4 x +

25 —

4 for y in Equation 1 and solve for x.

x2 + ( − 3 — 4 x +

25 —

4 ) 2 = 25

x2 + 9 —

16 x2 −

75 —

8 x +

625 —

16 = 25

16x2 + 9x2 − 150x + 625 = 400

25x2 − 150x + 225 = 0

x2 − 6x + 9 = 0

(x − 3)2 = 0

x − 3 = 0

x = 3

Substitute the value for x into Equation 2 and solve for y.

x = 3: y = − 3 — 4 (3) +

25 —

4

y = 4

The solution is (3, 4).

x

y

2

1

3

4

5

6

4 5 6321

(3, 4)


Chapter 5

51. Use the elimination method, then solve for y.

x2 + y2 = 1

−(x2 − 2 = 2y)

y2 + 2 = 1 − 2y

Solve for y.

y2 + 2 = 1 − 2y

y2 + 2y + 1 = 0

(y + 1)2 = 0

y + 1 = 0

y = −1

Substitute the value for y into Equation 1 and solve for x.

x2 + (−1)2 = 1

x

y2

−2

2−2(0, −1)

x2 + 1 = 1

x2 = 0

x = 0

The solution is (0, −1).

52. Use the elimination method, then solve for x.

x2 + y2 = 4

−(y2 = x + 2)

x2 = 4 − x − 2

Solve for x.

x2 = 4 − x − 2

x2 + x − 2 = 0

(x + 2)(x − 1) = 0

x + 2 = 0 or x − 1 = 0

x = −2 or x = 1

Substitute the values of x into Equation 2 and solve for y.

x = −2: y2 = −2 + 2 x = 1: y2 = 1 + 2

y2 = 0 y2 = 3

y = 0 y = ± √—

3

The solutions are (−2, 0), ( 1, √—

3 ) , and ( 1, − √—

3 ) .

x

y

3

4

−3

−4

431−3−4

(−2, 0)

(1, −√3)

(1, √3)

53. a. Dry asphalt: Wet asphalt:

45 = √—

30 ⋅ 0.75d 45 = √—

30 ⋅ 0.30d

45 = √—

22.5d 45 = √—

9d

452 = ( √—

22.5d ) 2 452 = ( √—

9d ) 2

2025 = 22.5d 2025 = 9d

90 = d 225 = d

Snow: Ice:

45 = √—

30 ⋅ 0.30d 45 = √—

30 ⋅ 0.15d

45 = √—

9d 45 = √—

4.5d

452 = ( √—

9d ) 2 452 = ( √—

4.5d ) 2

2025 = 9d 2025 = 4.5d

225 = d 450 = d

The greatest stopping distance is 450 feet on ice. On wet

asphalt and snow, the stopping distance is 225 feet. The

least stopping distance is 90 feet on dry asphalt.

b. Let s be 35 and f be 0.15 in the equation and solve for d.

35 = √—

30 ⋅ 0.15d

35 = √—

4.5d

(35)2 = ( √—

4.5d ) 2

1225 = 4.5d

272.2 ≈ d

The deer needs to be more than 272.2 feet away to stop

safely.

54. a. B = 0: 0 = 1.69 √—

s + 4.25 − 3.55

3.55 = 1.69 √—

s + 4.25

2.1 ≈ √—

s + 4.25

(2.1)2 ≈ ( √—

s + 4.25 ) 2

4.41 ≈ s + 4.25

0.16 ≈ s

The wind speed is about 0.16 mile per hour when B = 0.

B = 3: 3 = 1.69 √—

s + 4.25 − 3.55

6.55 = 1.69 √—

s + 4.25

3.876 ≈ √—

s + 4.25

(3.876)2 ≈ ( √—

s + 4.25 ) 2

15.02 ≈ s + 4.25

10.77 ≈ s

The wind speed is about 10.77 miles per hour when B = 3.

b. B = 12: 12 = 1.69 √—

s + 4.25 − 3.55

15.55 = 1.69 √—

s + 4.25

9.201 ≈ √—

s + 4.25

(9.201)2 ≈ ( √—

s + 4.25 ) 2

84.66 ≈ s + 4.25

80.41 ≈ s

From part (a), when B = 0, the wind speed is about

0.16 mile per hour and when B = 12 the wind speed is

about 80.41 miles per hour. So, the range of the wind

speeds is 0.16 ≤ s ≤ 80.41.


Chapter 5

55. Solve x − 4 = √—

2x .

x − 4 = √—

2x

(x − 4)2 = ( √—

2x ) 2

x2 − 8x + 16 = 2x

x2 − 10x + 16 = 0

(x − 8)(x − 2) = 0

x − 8 = 0 or x − 2 = 0

x = 8 or x = 2

Check:

8 − 4 ==?

√—

2 ⋅ 8 2 − 4 ==?

√—

2 ⋅ 2

4 ==?

√—

16 −2 ==?

√—

4

4 = 4 ✓ −2 ≠ 2 ✗


Solve x − 4 = − √—

2x .

x − 4 = − √—

2x .

(x − 4) = ( − √—

2x ) 2

x2 − 8x + 16 = 2x

x2 − 10x + 16 = 0

(x − 8)(x − 2) = 0

x − 8 = 0 or x − 2 = 0

x = 8 x = 2

Check:

8 − 4 =?

− √—

2(8) 2 − 4 =?

− √—

2(2)

4 =?

− √—

16 −2 =?

− √—

4

4 ≠ −4 ✗ −2 = −2 ✓


a. When solving the fi rst equation, the solution is x = 8

with x = 2 as an extraneous solution. When solving the

second equation, the solution is x = 2 with x = 8 as an

extraneous solution.

b.

x

y

2

1

3

4

5

6

7

8

−3

−4

−5

−6

−2

4 5 6 8 9 1031−1−2−3

(2, 0) (8, 0)

y = x − 4 − √2x

y = x − 4 + √2x

56. The friend is incorrect; Sample answer: It is possible to have

two extraneous solutions, for example, when the radical is a

fourth root.

57. You know there is no real solution because the square root of

a quantity cannot be negative.

58. The solution of the equation is x = 5. The solution is the

x-value of the point of intersection of the graphs.

59. If the price is raised, then the demand on the number of units

will decrease.

60. The area of the circle must be at least 6P. For a circle with

radius r, this means πr2 ≥ 6P. Solve the inequality for r.

πr2 ≥ 6P

r2 ≥ 6P

— π

r ≥ √—

6P

— π

61. Set r equal to 3 in the equation and solve for s.

3 = 1 —

2 √—

S —

π

6 = √—

S —

π

62 = ( √—

S —

π ) 2

36 = S —

π

36π = S

113.1 ≈ S

The surface area of the Moetaki Boulder is about

113.1 square feet.

62. Let ℓ be 5, b1 be 2, and b2 be 4 in the equation and solve

for h.

5 = √——

h2 + 1 —

4 (4 − 2)2

5 = √— h2 + 1 —

4 (2)2

5 = √—

h2 + 1

52 = ( √—

h2 + 1 ) 2

25 = h2 + 1

24 = h2

± √—

24 = h

±4.9 ≈ h

Reject the negative solution because height cannot be negative.

So, the height of the truncated pyramid is about 4.9 feet.

63. a. r = √— kt —

π(h0 − h) b. h = h0 −

kt —

πr2

r2 = kt —

π(h0 − h) = 6.5 −

0.04(45) —

π(0.875)2

h0 − h = kt —

πr2 ≈ 5.75

h = h0 − kt —

πr2 After 45 minutes, the

height of the candle is

about 5.75 inches.


Chapter 5


64. (x3 − 2x2 + 3x + 1) + (x4 − 7x)

= x4 + x3 − 2x2 + (3 − 7)x + 1

= x4 + x3 − 2x2 − 4x + 1

65. (2x5 + x4 − 4x2) − (x5 − 3) = (2 − 1)x5 + x4 − 4x2 + 3

= x5 + x4 − 4x2 + 3

66. (x3 + 2x2 + 1)(x2 + 5)

= (x3 + 2x2 + 1)(x2) + (x3 + 2x2 + 1)(5)

= x5 + 2x4 + x2 + 5x3 + 10x2 + 5

= x5 + 2x4 + 5x3 + 11x2 + 5

x3 + 11x − 8

67. x + 2 ) ‾‾‾ x4 + 2x3 + 11x2 + 14x − 16

x4 + 2x3

0x3 + 11x2 + 14x

11x2 + 22x

−8x − 16

−8x − 16

0

x4 + 2x3 + 11x2 + 14x − 16

——— x + 2

= x3 + 11x − 8

68. g(x) = f (−x) + 4

= [ (−x)3 − 4(−x)2 + 6 ] + 4

= −x3 − 4x2 + 10

The graph of g is a refl ection in the y-axis followed by a

translation 4 units up of the graph of f.

69. g(x) = 1 —

2 f (x) − 3

= 1 —

2 (x3 − 4x2 + 6) − 3

= 1 —

2 x3 − 2x2

The graph of g is a vertical shrink by a factor of 1 —

2 followed

by a translation 3 units down of the graph of f.

70. g(x) = −f (x − 1) + 6

= − [ (x − 1)3 − 4(x − 1)2 + 6 ] + 6

= −(x − 1)3 + 4(x − 1)2

The graph of g is a refl ection in the x-axis followed by a

translation 1 unit right and 6 units up of the graph of f.


1. a.

x

y

2

1

3

4

5

6

7

8

−3

−4

−2

4 5 6321−2−1−3−4−5−6

f + g

b.

x

y

2

1

3

4

−3

−4

−5

−6

−7

−8

−2

4 5 63−3−4−5−6

f + g

2. Measure to fi nd y-values of one function and add to the

corresponding y-values of the other function.

3. a. The equation for the function f in Exploration 1(a) is

f(x) = 0.25(x − 2)2 and g is g(x) = x − 1. The sum

of the functions is

f(x) + g(x) = 0.25(x − 2)2 + (x − 1)

= 0.25(x2 − 4x + 4) + x − 1

= 0.25x2 − x + 1 + x − 1

= 0.25x2.

b. The equation for the function f in Exploration 1(b) is

f(x) = −0.25(x − 2)2 and g is g(x) = −x + 1. The sum

of the functions is

f(x) + g(x) = −0.25(x − 2)2 + (−x + 1)

= −0.25(x2 − 4x + 4) − x + 1

= −0.25x2 + x − 1 − x + 1

= −0.25x2.


Chapter 5


1. (f + g)(x) = f(x) + g(x)

= −2x2/3 + 7x2/3

= 5x2/3

The functions f and g each have the same domain: all real

numbers. So, the domain of f + g is all real numbers. When

x = 8, the value of the sum is

(f + g)(8) = 5(8)2/3

= 5(4)

= 20.

( f − g)(x) = f(x) − g(x)

= −2x2/3 − 7x2/3

= −9x2/3


numbers. So, the domain of f − g is all real numbers. When

x = 8, the value of the difference is

( f − g)(8) = −9(8)2/3

= −9(4)

= −36.

2. ( fg)(x) = f(x)g(x)

= (3x)(x1/5)

= 3x6/5


numbers. So, the domain of fg is all real numbers. When

x = 32, the value of the product is

( fg)(32) = 3(32)6/5 = 3(64) = 192.

( f — g ) (x) =

3x —

x1/5 = 3x4/5


numbers. So, the domain of f —

g is all real numbers except 0.

When x = 32, the value of the quotient is

( f — g ) (32) = 3(32)4/5 = 3(16) = 48.

3. Enter the functions y1 = 8x and y2 = 2x5/6 in a graphing

calculator. On the home screen, enter y1(5) + y2(5). The fi rst

entry on the screen shows that y1(5) + y2(5) ≈ 47.65, so ( f + g)(5) ≈ 47.65. Enter the other operations.

( f − g)(5) ≈ 32.35 ( fg)(5) ≈ 305.89

( f — g ) (5) ≈ 5.23

4. g(3) = √—

9 − 32 = 0, so 3 is not in the domain of f —

g .

5. (rs)(1.7 × 105) = (1.446 × 109)(1.7 × 105)−0.05

≈ 791,855,335

The white rhino’s heart beats about 791,855,335 times.



1. You can add, subtract, multiply, or divide f and g.

2. Any x-values not in the domains of both functions and any

x-values that result in the denominator being equal to 0 are

not included in the domain of the quotient of two functions.

Monitoring Progress and Modeling With Mathematics

3. ( f + g)(x) = f(x) + g(x) = −5 4 √—

x + 19 4 √—

x = 14 4 √—

x

The functions f and g each have the same domain: x ≥ 0. So,

the domain of f + g is x ≥ 0. When x = 16, the value of the

sum is

(f + g)(16) = 14 4 √—

16 = 14(2) = 28.

( f − g)(x) = f(x) − g(x)

= −5 4 √—

x − 19 4 √—

x = −24 4 √—

x


the domain of f − g is x ≥ 0. When x = 16, the value of the

difference is

( f − g)(16) = −24 4 √—

16 = −24(2) = −48.

4. ( f + g)(x) = f(x) + g(x) = 3 √—

2x − 11 3 √—

2x = −10 3 √—

2x



x = −4, the value of the sum is

(f + g)(−4) = −10 3 √—

2(−4)

= −10 3 √—

(−8) = (−10)(−2) = 20.

( f − g)(x) = f(x) − g(x)

= 3 √—

2x + 11 3 √—

2x = 12 3 √—

2x



x = −4, the value of the difference is

( f − g)(−4) = 12 3 √—

2(−4) = 12(−2) = −24.

5. ( f + g)(x) = f(x) + g(x) = (6x − 4x2 − 7x3) + (9x2 − 5x)

= −7x5 + 5x2 + x



x = −1, the value of the sum is

( f + g)(−1) = −7(−1)3 + 5(−1)2 + (−1) = 11.

( f − g)(x) = f(x) − g(x) = (6x − 4x2− 7x3) − (9x2 − 5x)

= −7x3 − 13x2 + 11x



x = −1, the value of the difference is

( f − g)(−1) = −7(−1)3 − 13(−1)2 + 11(−1) = −17.


Chapter 5

6. ( f + g)(x) = f(x) + g(x)

= (11x + 2x2) + (−7x − 3x2 + 4)

= −x2 + 4x + 4



x = 2, the value of the sum is

( f + g)(2) = −(2)2 + 4(2) + 4 = 8.

( f − g)(x) = f(x) − g(x)

= (11x + 2x2) − (−7x − 3x2 + 4)

= 5x2 + 18x − 4



x = 2, the value of the difference is

( f − g)(2) = 5(2)2 + 18(2) − 4 = 52.

7. ( fg)(x) = (2x3) ( 3 √—

x ) = 2x10/3



x = −27, the value of the product is

( fg)(−27) = 2(−27)10/3 = 2(59,049) = 118,098.

( f — g ) (x) =

2x3

— 3 √

— x = 2x8/3



g is x ≠ 0. When

x = −27, the value of the quotient is

( f — g ) (−27) = 2(−27)8/3 = 2(6561) = 13,122.

8. ( fg)(x) = (x4)(3 √—

x ) = 3x9/2

The domain of f is all real numbers and the domain of g is

x ≥ 0. So, the domain of fg is x ≥ 0. When x = 4, the value of

the product is ( fg)(4) = 3(4)9/2 = 3(512) = 1536.

( f — g ) (x) =

x4

— 3 √

— x =

x7/2

— 3


x ≥ 0. So, the domain of f —

g is x > 0. When x = 4, the value

of the quotient is ( f — g ) (4) =

47/2

— 3 =

128 —

3 .

9. ( fg)(x) = (4x)(9x1/2) = 36x3/2


x ≥ 0. So, the domain of fg is x ≥ 0. When x = 9, the value of

the product is ( fg)(9) = 36(9)3/2 = 36(27) = 972.

( f — g ) (x) =

4x —

9x1/2 =

4 —

9 x1/2


x ≥ 0. So, the domain of f —

g is x > 0. When x = 9, the value

of the quotient is ( f — g ) (9) =

4 —

9 (9)1/2 =

4 —

9 (3) =

4 —

3 .

10. ( fg)(x) = (11x3)(7x7/3) = 77x16/3



x = −8, the value of the product is

(fg)(−8) = 77(−8)16/3 = 77(65,536) = 5,046,272.

( f — g ) (x) =

11x3

— 7x7/3

= 11

— 7 x2/3



g is x ≠ 0. When x = −8, the

value of the quotient is ( f — g ) (−8) =

11 —

7 (−8)2/3 =

11 —

7 (4) =

44 —

7 .

11. ( fg)(x) = f (x)g(x) = (7x3/2)(−14x1/3) = −98x11/6

The domain of f is x ≥ 0 and the domain of g is all real

numbers. So, the domain of fg is x ≥ 0. When


(fg)(64) = −98(64)11/6 = −98(2048) = −200,704.

( f — g ) (x) =

f (x) —

g(x) =

7x3/2

— −14x1/3

= − 1 —

2 x7/6



g is x > 0. When x = 64, the

value of the quotient is

( f — g ) (64) = −

1 —

2 (64)7/6 = −

1 —

2 (128) = −64.

12. ( fg)(x) = (4x5/4)(2x1/2) = 8x7/4

The functions f and g each have the same domain: x ≥ 0.

So, the domain of fg is x ≥ 0. When x = 16, the value of the

product is (fg)(16) = 8(16)7/4 = 8(128) = 1024.

( f — g ) (x) =

f (x) —

g(x) =

4x5/4

— 2x1/2

= 2x3/4


the domain of f —

g is x > 0. When x = 16, the value of the

quotient is ( f — g ) (16) = 2(16)3/4 = 2(8) = 16.

13. Enter the functions y1 = 4x4 and y2 = 24x1/3 in a graphing


entry on the screen shows that y1(5) + y2(5) ≈ 2541.04. So,

( f + g)(5) ≈ 2541.04. Enter the other operations.

(f − g)(5) ≈ 2458.96

(fg)(5) ≈ 102,598.56

( f — g ) (5) ≈ 60.92

14. Enter the functions y1 = 7x5/3 and y2 = 49x2/3 in a graphing


entry on the screen shows that y1(5) + y2(5) ≈ 245.62. So, ( f + g)(5) ≈ 245.62. Enter the other operations.

( f − g)(5) ≈ −40.94 (fg)(5) ≈ 14,663.04 ( f — g ) (5) ≈ 0.71


Chapter 5

15. Enter the functions y1 = −2x1/3 and y2 = 5x1/2 in a graphing


entry on the screen shows y1(5) + y2(5) ≈ 7.76. So,


( f − g)(5) ≈ −14.60 (fg)(5) ≈−38.24 ( f — g ) (5) ≈ −0.31

16. Enter the functions y1 = 4x1/2 and y2 = 6x3/4 in a graphing


entry on the screen shows that y1(5) + y2(5) ≈ 29.01. So,


( f − g)(5) ≈ −11.12 (fg)(5) ≈ 179.44 ( f — g

) (5) ≈ 0.45

17. Because the functions have an even index, the domain is

restricted; The domain of ( fg)(x) is x ≥ 0.

18. The domain is incorrect; The domain of ( f — g ) (x) is all real

numbers except x = 2 and x = −2.

19. a. (F + M)(t) = F(t) + M(t)

= (−0.007t2 + 0.10t + 3.7)

+ (0.0001t3 − 0.009t2 + 0.11t + 3.7)

= 0.0001t3 − 0.016t2 + 0.21t + 7.4

b. The sum represents the total number of employees (in

millions) from the ages of 16 to 19 in the United States

from 1990 to 2010.

20. a. (W − F)(t) = W(t) − F(t)

= (−5.8333t3 + 17.43t2 + 509.1t + 11,496)

− (12.5t3 − 60.29t2 + 136.6t + 4881)

= −18.3333t3 + 77.72t2 + 372.5t + 6615

b. The difference represents the number of cruise ship

departures (in thousands) from around the world except

for Florida.

21. Your friend is correct. The addition and multiplication

of functions are commutative because addition and

multiplication of real numbers is commutative, and the order

in which they appear does not matter.

22. B; A; The y-intercept in A is less than in B.

23. ( f + g)(3) = f(3) + g(3) = 10 − 31 = −21

( f − g)(1) = f(1) − g(1) = −4 − (−3) = −1

( fg)(2) = (0)(−13) = 0

( f — g ) (0) =

−2 —

−1 = 2

24. It is possible to write a sum of two functions containing a

radical, but whose product does not. This can happen when,

for example, both functions are radical functions with the

same radicand. An example is f (x) = √—

x and g(x) = 2 √—

x .

25. r(x) = (area of square) − (area of triangle)

= x2 − 1 —

2 x ⋅ x

= x2 − 1 —

2 x2

= 1 —

2 x2

26. r(w) = 1.1w0.734

—— b(w) − d(w)

= 1.1w0.734

—— 0.007w − 0.002w

= 1.1w0.734

— 0.005w

= 220w0.734

— w

When w = 6.5: r(6.5) = 220(6.5)0.734

—— 6.5

≈ 133.7

When w = 300: r(300) = 220(300)0.734

—— 300

≈ 48.3

When w = 70,000: r(70,000) = 220(70,000)0.734

—— 70,000

≈ 11.3

27. a. r(x) = 20 − x

— 6.4

, s(x) = √—

x2 + 144 —

0.9

b. t(x) = r(x) + s(x) = 20 − x

— 6.4

+ √—

x2 + 144 —

0.9

c.

200

0

30

MinimumX=1.7044372Y=16.325839

From the graph, x ≈ 1.7 minimizes t. If Elvis runs along

the shore until he is about 1.7 meters from point C then

swims to point B, the time taken to get there will be a

minimum.


28. 3xn − 9 = 6y 29. 5z = 7n + 8nz

3xn = 6y + 9 5z = n(7 + 8z)

n = 6y + 9

— 3x

5z —

7 + 8z = n

n = 2y + 3

— x

30. 3nb = 5n − 6z 31. 3 + 4n — n = 7b

3nb − 5n = −6z 3 + 4n = 7bn

n(3b − 5) = −6z 3 = 7bn − 4n

n = −6z

— 3b − 5

= 6z —

5 − 3b 3 = n(7b − 4)

3 —

7b − 4 = n

32. The relation is a function because every input has exactly

one output.

33. The relation is not a function because −1 has two outputs.


Chapter 5

34. The relation is a function because every input has exactly

one output.

35. The relation is not a function because 2 has two outputs.


1. a. fy

3

4

5

−3

−4

−5

−6

−2

4 5 63−2−3−4−5−6

g

The graph of g is a refl ection in the line y = x of the graph

of f.

b.

x

y

2

3

4

5

6

−3

−4

−5

−6

−2

4 5 632−2−3−4−5−6

g

f


of f.

c.

x

y

2

1

3

4

5

6

7

8

9

10

11

12

4 5 6 7 8 9 10 11 12321

g

f


of f.

d.

x

y

2

1

3

4

5

6

−3

−4

−5

−6

−2

5 6321−2−3−4−5−6

g

f


of f.

2. a.

x

y

4

8

−8

84−8

g

f

The graph of g is a refl ection in the line y = x.

b.

x

y

4

8

−4

−8

84−4−8

g

f


c.

x

y8

4

−8

−4

84−8

g

f


d.

x

y8

4

−8

−4

84−2−4

g

f


3. To graph the inverse of a function, graph the original

function and then refl ect the graph in the line y = x to form

the graph of the inverse function.


Chapter 5

4. The operations in one equation are the inverses of the

operations in the other. So, g(x) = x + 3

— 2 .

x

y8

−8

84−8

f(x) = 2x − 3

g(x) = x + 32


1. y = x − 2

y + 2 = x

Find the input when y = 2.

x = 2 + 2

= 4

So, the input is 4 when the output is 2.

2. y = 2x2

y — 2 = x2

± √—

y —

2 = x


x = ± √—

2 —

2

= ± √—

1

= ±1

So, the input is −1 or 1 when the output is 2.

3. y = −x3 + 3

y − 3 = −x3

−y + 3 = x3

3 √—

−y + 3 = x


x = 3 √—

−2 + 3 = 3 √—

1 = 1

So, the input is 1 when the output is 2.

4. Method 1 Use inverse operations in the reverse order.

f(x) = 2x

To fi nd the inverse, apply inverse operations in

the reverse order.

g(x) = x —

2

The inverse of f is g(x) = x —

2 .

Method 2 Set y equal to f(x). Switch the roles of x and y

and solve for y.

y = 2x

x = 2y

x —

2 = y

The inverse of f is g(x) = x —

2 .

x

y

2

1

3

−3

−2

321−2−1−3

g

f


f(x) = −x + 1


the reverse order.

g(x) = −(x − 1)

The inverse of f is g(x) = −(x − 1) or

g(x) = −x + 1.


and solve for y.

y = −x + 1

x = −y + 1

x − 1 = −y

−(x + 1) = y

−x + 1 = y

The inverse of f is g(x) = −x + 1.

x

y

2

1

3

−3

−2

31−2−1−3

f

g


Chapter 5


f(x) = 1 —

3 x − 2


the reverse order.

g(x) = 3(x + 2)

The inverse of f is g(x) = 3(x + 2) or

g(x) = 3x + 6.


and solve for y.

y = 1 —

3 x − 2

x = 1 —

3 y − 2

x + 2 = 1 —

3 y

3(x + 2) = y

3x + 6 = y

The inverse of f is g(x) = 3x + 6.

x

y

2

1

3

−3

−4

−5

−6

−7

−8

321−1−3−4−5−6−7−8

g

f

7. f (x) = −x2, x ≤ 0

y = −x2

x = −y2

−x = y2

± √—

−x = y

The domain of f is restricted to nonpositive values of x. So,

the range of the inverse must also be restricted to nonpositive

values. So, the inverse of f is g(x) = − √—

−x .

x

y1

−3

−4

−5

−6

−2

1−2−3−4−5−6

g

f

8. f (x) = −x3 + 4

y = −x3 + 4

x = −y3 + 4

x − 4 = −y3

−x + 4 = y3

3 √—

−x + y = y

So the inverse of f is

x

y

1

3

5

6

−3

−2

5 631−2−1−3g

fg(x) = 3 √—

−x + 4 .

9. f (x) = √—

x + 2

y = √—

x + 2

x = √—

y + 2

x2 = y + 2

x2 − 2 = y

Because the range of f is y ≥ 0, the domain of the

inverse must be restricted to x ≥ 0. So, the inverse

of f is g(x) = x2 − 2, x ≥ 0.

x

y

2

3

4

5

6

−3

−2

4 5 632−2−1−3

g

f

10. Step 1 Show that f (g(x)) = x.

f (g(x)) = f (x − 5)

= (x − 5) + 5

= x − 5 + 5

= x ✓

Step 2 Show that g( f (x)) = x.

g( f (x)) = g(x + 5)

= (x + 5) − 5

= x + 5 − 5

= x ✓

The functions are inverse functions.


Chapter 5


f (g(x)) = f ( 3 √—

2x )

= 8 ( 3 √—

2x ) 3

= 8(2x)

= 16x ✗

The functions are not inverse functions.

12. Step 1 Find the inverse of the function.

d = 4.9t2

d —

4.9 = t2

√—

d —

4.9 = t

Step 2 Evaluate the inverse when d = 50.

t = √—

50

— 4.9

≈ 3.19

It takes an object about 3.19 seconds to fall 50 meters.



1. Inverse functions are functions that undo each other.

2. The horizontal line test can be used to determine whether the

inverse of a function is a function.

3. Functions f and g are inverses of each other provided that

f (g(x)) = x and g( f (x)) = x.

4. “Write an equation that represents a refl ection of the graph

of f(x) = 5x − 2 in the x-axis” is different from the others.

The result of this is y = −5x + 2; The other answer is

y = 1 —

5 (x + 2).


5. y = 3x + 5

y − 5 = 3x

y − 5

— 3 = x

Find the input when y = −3.

x = −3 − 5

— 3 = −

8 —

3

So, the input is − 8 — 3 when the output is −3.

6. y = −7x − 2

y + 2 = −7x

y + 2

— −7

= x


x = −3 + 2

— −7

= −1

— −7

= 1 —

7

So, the input is 1 —

7 when the output is −3.

7. y = 1 —

2 x − 3

y + 3 = 1 —

2 x

2(y + 3) = x

2y + 6 = x


x = 2(−3 + 3)

= 2(0)

= 0

So, the input is 0 when the output is −3.

8. y = − 2 — 3 x + 1

y − 1 = − 2 — 3 x

− 3 — 2 (y − 1) = x

− 3y − 3

— 2 = x


x = − 3 — 2 (−3 − 1)

= − 3 — 2 (−4)

= 6

So, the input is 6 when the output is −3.

9. y = 3x3

y — 3 = x3

3 √—

y —

3 = x


x = 3 √—

−3

— 3

= 3 √—

−1

= −1

So, the input is −1 when the output is −3.

10. y = 2x4 − 5

y + 5 = 2x4

y + 5

— 2 = x4

± 4 √—

y + 5

— 2 = x


x = ± √— −3 + 5

— 2

= ± √—

2 —

2

= ± √—

1

= ±1

So, the input is −1 or 1 when the output is −3.


Chapter 5

11. y = (x − 2)2 − 7

y + 7 = (x − 2)2

± √—

y + 7 = x − 2

± √—

y + 7 + 2 = x


x = ± √—

−3 + 7 + 2

= ± √—

4 + 2

= ±2 + 2

= 0 or 4

So, the input is 0 or 4 when the output is −3.

12. y = (x − 5)3 − 1

y + 1 = (x − 5)3

3 √—

y + 1 = x − 5

3 √—

y + 1 + 5 = x


x = 3 √—

−3 + 1 + 5

= 3 √—

−2 + 5

So, the input is 3 √—

−2 + 5 when the output is −3.

13. Method 1 Use inverse operations in reverse order.

f(x) = 6x


the reverse order.

g(x) = 1 — 6 x

The inverse of f is g(x) = 1 —

6 x.


and solve for y.

y = 6x

x = 6y

1 —

6 x = y


6 x.

x

y

2

1

321−1

g

f


f(x) = −3x


the reverse order.

g(x) = − 1 — 3 x

The inverse of f is g(x) = − 1 — 3 x.


and solve for y.

y = −3x

x = −3y

− 1 — 3 x = y

The inverse of f is g(x) = − 1 — 3 x.

x

y3

−3

−2

3−2−1−3g

f


f(x) = −2x + 5


the reverse order.

g(x) = − 1 — 2 (x − 5)

The inverse of f is g(x) = − 1 — 2 (x − 5) or

g(x) = − 1 — 2 x +

5 —

2 .


and solve for y.

y = −2x + 5

x = −2y + 5

x − 5 = −2y

− 1 — 2 (x − 5) = y

The inverse of f is g(x) = − 1 — 2 (x − 5) or

g(x) = − 1 — 2 x +

5 —

2 .

x

y

2

1

4

5

7

8

−2

4 5 7 821−2−1g

f


f(x) = 6x − 3


the reverse order.

g(x) = 1 —

6 (x + 3)


6 (x + 3) or

g(x) = x + 3

— 6 .


Chapter 5


and solve for y.

y = 6x − 3

x = 6y − 3

x + 3 = 6y

x + 3

— 6 = y

The inverse of f is g(x) = x + 3

— 6 .

y

2

1

3

4

5

−3

−2

4 5321−2−1−3

g

f


f(x) = − 1 — 2 x + 4


the reverse order.

g(x) = −2(x − 4)

The inverse of f is g(x) = −2(x − 4) or

g(x) = −2x + 8.


and solve for y.

y = − 1 — 2 x + 4

x = − 1 — 2 x + 4

x − 4 = − 1 — 2 y

−2(x − 4) = y

The inverse of f is g(x) = −2(x − 4) or

g(x) = −2x + 8.

x

y

2

1

3

5

6

7

−2

5 6 7 8321−2−1

g

f


f(x) = 1 —

3 x − 1


the reverse order.

g(x) = 3(x + 1)

The inverse of f is g(x) = 3(x + 1), or

g(x) = 3x + 3.


and solve for y.

y = 1 —

3 x − 1

x = 1 —

3 y − 1

x + 1 = 1 —

3 y

3(x + 1) = y

The inverse of f is g(x) = 3(x + 1), or

g(x) = 3x + 3.

x

y

3

4

−3

−4

−5

−2

4 53−2−3−4−5

g

f


f(x) = 2 —

3 x −

1 —

3


reverse order.

g(x) = 3 —

2 ( x +

1 — 3 )


2 ( x +

1 — 3 ) , or

g(x) = 3 —

2 x +

1 —

2 .


and solve for y.

y = 2 —

3 x −

1 —

3

x = 2 —

3 y −

1 —

3

x + 1 —

3 =

2 —

3 y

3 —

2 ( x +

1 —

3 ) = y


2 ( x +

1 —

3 ) , or

g(x) = 3 —

2 x +

1 —

2 .

x

y

2

1

3

4

−3

−4

−2

4321−2−1−3−4

g

f


Chapter 5


f(x) = − 4 — 5 x +

1 —

5


reverse order.

g(x) = − 5 — 4 ( x −

1 —

5 )

The inverse of f is g(x) = − 5 — 4 ( x −

1 —

5 ) , or

g(x) = − 5 — 4 x +

1 —

4 .


and solve for y.

y = − 4 — 5 x +

1 —

5

x = − 4 — 5 y +

1 —

5

x − 1 —

5 = −

4 — 5 y

− 5 — 4 ( x −

1 —

5 ) = y

The inverse of f is g(x) = − 5 — 4 ( x −

1 —

5 ) , or

g(x) = − 5 — 4 x +

1 —

4 .

x

y

2

3

−3

−2

32−2−1−3

g

f

21. Switching the variables:

y = −3x + 4

x = −3y + 4

x − 4 = −3y

− 1 — 3 (x − 4) = y

The inverse function of f is g(x) = − 1 — 3 (x − 4). Reversing the

operations:

g(x) = − 1 — 3 (x − 4)

The inverse function of f is g(x) = − 1 — 3 (x − 4).

Sample answer: The method of using inverse operations in

reverse order is preferred because there is no computation.

22. a. The functions are inverses because the coordinates switch

roles.

b. The functions are not inverses because the coordinates did

not switch roles.

c. The functions are not inverses because the coordinates did

not switch roles.

23. f(x) = 4x2, x ≤ 0

y = 4x2

x = 4y2

x — 4 = y2

± √

— x —

2 = y

The domain of f is restricted

x

y

3

4

5

6

−3

−2

4 5 6−2−1−3g

f

to nonpositive values of x. So,

the range of the inverse must

also be restricted to nonpositive

values. So, the inverse of f is

g(x) = − √

— x —

2 .

24. f(x) = 9x2, x ≤ 0

y = 9x2

x = 9y2

x — 9 = y2

± √

— x —

3 = y


x

y

−2

−2−1g

f

to nonpositive values of x. So,


also be restricted to nonpositive

values. So, the inverse of f is

g(x) = − √

— x —

3 .

25. f(x) = (x − 3)3

y = (x − 3)3

x = (y − 3)3

3 √—

x = y − 3

3 √—

x + 3 = y

The inverse of f is g(x) = 3 √—

x + 3.

x

y

1

3

4

5

6

−3

−4

−2

4 5 631−2−1−3−4

g

f


Chapter 5

26. f(x) = (x + 4)3

y = (x + 4)3

x = (y + 4)3

3 √—

x = y + 4

3 √—

x − 4 = y


x − 4.

x

y

2

1

3

4

−3

−4

−5

−6

−7

−8

−9

−10

−2

4321−2−1−3−4−5−6−7−8−9−10

g

f

27. f(x) = 2x4, x ≥ 0

y = 2x4

x = 2y4

x — 2 = y4

± 4 √—

x —

2 = y


x

y

2

1

3

4

5

6

4 5 6321−1

g

fto nonnegative values of x.

So, the range of the inverse

must also be restricted to

nonnegative values. So, the

inverse of f is

g(x) = 4 √—

x —

2 .

28. f(x) = −x6, x ≥ 0

y = −x6

x = −y6

−x = y6

± 6 √—

−x = y


x

y2

−3

−4

−5

−2

2−2−1−3−4−5

g

f

to nonnegative values of x. So,


also be restricted to nonnegative

values. So, the inverse of f isg(x) = 6 √

— −x .

29. When switching x and y, the negative should not be switched

with the variables.

f(x) = −x + 3

y = −x + 3

x = −y + 3

x − 3 = −y

−x + 3 = y

30. The inverse should only be y = √—

7x because the domain of f is x ≥ 0.

f(x) = 1 —

7 x2, x > 0

y = 1 —

7 x2

x = 1 —

7 y2

7x = y2

√—

7x = y

31. f does not have an inverse function because the graph of f does not pass the horizontal line test.



34. f does have an inverse function because the graph of f passes

the horizontal line test.

35. Graph the function.

Notice that no horizontal line

intersects the graph more than

once. So, the inverse of f is a

function. Find the inverse.

y = x3 − 1

x

y

20

10

−20

−10

42−2−4 x = y3 − 1

x + 1 = y3

3 √—

x + 1 = y

So, the inverse of f is g(x) =

3 √—

x + 1 .


Notice that no horizontal

line intersects the graph

more than once. So, the

inverse of f is a function.

Find the inverse.

y = −x3 + 3 x

y

20

10

−20

−10

4−2−4

x = −y3 + 3

x − 3 = −y3

−x + 3 = y3

3 √—

−x + 3 = y

So, the inverse of f is g(x) = 3 √—

−x + 3 .


Chapter 5



x

y

4

−2

42−2−4


once. So, the inverse of f is a

function. Find the inverse.

y = √—

x + 4

x = √—

y + 4

x2 = y + 4

x2 − 4 = y



of f is g(x) = x2 − 4, x ≥ 0.



x

y

4

6

2

8 124


once. So, the inverse of f isa function. Find the inverse.

y = √—

x − 6

x = √—

y − 6

x2 = y − 6

x2 + 6 = y



of f is g(x) = x2 + 6, x ≥ 0.



x

y4

2

−4

−2

8 124−4


once. So, the inverse of fis a function. Find the inverse.

y = 2 3 √—

x − 5

x = 2 3 √—

y − 5

x — 2 =

3 √—

y − 5

x3

— 8 = y − 5

x3 —

8 + 5 = y

So, the inverse of f is g(x) = x3

— 8 + 5.


Notice that a horizontal line

x

y

8

4

42−2−4

12can intersect the graph more

than once. So, the inverse is

not a function. Find the inverse.

y = 2x2 − 5

x = 2y2 − 5

x + 5 = 2y2

x − 5

— 2 = y2

± √— x − 5

— 2 = y


x

y

20

10

2−2

30

Notice that a horizontal line can intersect the graph more

than once. So, the inverse is not a function. Find the inverse.

y = x4 + 2

x = y4 + 2

x − 2 = y4

± 4 √—

x − 2 = y


x

y

20

10

−20

42−2−4

Notice that no horizontal line intersects the graph more than

once. So, the inverse of f is a function. Find the inverse.

y = 2x3 − 5

x = 2y3 − 5

x + 5 = 2y3

x + 5

— 2 = y3

3 √—

x + 5

— 2 = y


x + 5

— 2 .




once. So, the inverse of fis a function. Find the inverse.

y = 3 3 √—

x + 1

x = 3 3 √—

y + 1

x — 3 =

3 √—

y + 1

x

y8

4

−8

−4

42−2−4

( x — 3 ) 3 = y + 1

x3 —

27 − 1 = y

So, the inverse of f is g(x) = x3

— 27

− 1.


Chapter 5


x

y4

2

−4

−2

42−4



y = − 3 √— 2x + 4

— 3

x = − 3 √— 2y + 4

— 3

−x = 3 √— 2y + 4

— 3

−x3 = 2y + 4

— 3

−3x3 = 2y + 4

−3x3 − 4 = 2y

−3x3 − 4

— 2 = y

So, the inverse of f is g(x) = −3x3 − 4

— 2 .


x

y

20

10

−20

−10

42−2−4


once. So, the inverse of f is a function.

Find the inverse.

y = 1 —

2 x5

x = 1 —

2 y5

2x = y5

5 √—

2x = y


2x .


x

y

2

−4

−2

4−2−4



y = −3 √— 4x − 7

— 3

x = −3 √— 4y − 7

— 3

x —

−3 = √—

4y − 7 —

3

( x —

−3 ) 2 =

4y − 7 —

3

3 ( x2 —

9 ) = 4y − 7

x2

— 3 + 7 = 4y

x2

— 12

+ 7 —

4 = y

x2 + 21

— 12

= y

Because the range of f is y ≤ 0, the domain of the inverse

must be restricted to x ≤ 0. So, the inverse of f is

g(x) = x2 + 21

— 12

, x ≤ 0.

47. B; The function has a reciprocal slope and positive y-intercept.

48. C; The inverse operations are in the reverse order.


f (g(x)) = f ( x — 2 + 9 )

= 2 ( x — 2 + 9 ) − 9

= x + 18 − 9

= x + 9 ✗

The functions are not inverses.


Chapter 5


f (g(x)) = f (4x + 3)

= (4x + 3) − 3

—— 4

= 4x + 3 − 3

— 4

= 4x

— 4

= x ✓

Step 2 Show that g( f(x)) = x.

g( f(x)) = g ( x − 3 —

4 )

= 4 ( x − 3 —

4 ) + 3

= x − 3 + 3

= x ✓

The functions are inverses.


f (g(x)) = f(5x5 − 9)

= 5 √——

(5x5 − 9) + 9

—— 5

= 5 √——

5x5 − 9 + 9

—— 5

= 5 √—

5x5

— 5

= 5 √—

x5

= x ✓

Step 2 Show that g( f(x)) = x.

g( f(x)) = g ( 5 √—

x + 9

— 5 )

= 5 ( 5 √—

x + 9

— 5 ) 5 − 9

= 5 ( x + 9 —

5 ) − 9

= x + 9 − 9

= x ✓



f (g(x)) = f ( ( x + 4 —

7 )

3/2

) = 7 [ ( x + 4

— 7 )

3/2

] 3/2

− 4

= 7 ( x + 4 —

7 )

9/2

− 4 ✗

The functions are not inverses.


v = 1.34 √—

ℓ

v —

1.34 = √

— ℓ

( v —

1.34 )

2 = ℓ

Step 2 Evaluate the inverse when v = 7.5.

ℓ= ( 7.5 —

1.34 ) 2 ≈ 31.3

The hull length is about 31.3 feet.


R = 3 —

8 L − 5

R + 5 = 3 —

8 L

8 —

3 (R + 5) = L

Step 2 Evaluate the inverse when R = 19.

L = 8 —

3 (19 + 5)

= 8 —

3 (24)

= 64

The length of the stretched band is 64 inches.

55. B; When you refl ect the graph in the line y = x, you get the

graph shown in B.

56. C; When you refl ect the graph in the line y = x, you get the

graph shown in C.

57. A; When you refl ect the graph in the line y = x, you get the

graph shown in A.

58. D; When you refl ect the graph in the line y = x, you get the

graph shown in D.

59. If the number is n, then the fi nal number is given by

r = 2n2 + 3. The inverse function is n = √— r − 3

— 2 . So,

the original number was n = √— 53 − 3

— 2 = 5.

60. Your friend is incorrect. In order to guarantee that a quadratic

function has an inverse function, restrict the domain to

values greater than or equal to the x-coordinate of the

vertex. For example, y = (x − 1)2, x ≥ 0 does not have an

inverse function but y = (x − 1)2, x ≥ 1 does have an inverse

function.

61. a. First, fi nd the inverse function.

ℓ= 0.5w + 3

ℓ− 3 = 0.5w

2(ℓ− 3) = w

The function represents the weight of an object on a

stretched spring of length ℓ. b. Letℓbe 5.5 in the inverse from part (a).

w = 2(5.5 − 3) = 5

The weight is 5 pounds.


Chapter 5

c. Step 1 Show that ℓ(w(ℓ)) =ℓ.

ℓ(w(ℓ)) =ℓ(2(ℓ − 3))

= 0.5 [ 2(ℓ − 3) ] + 3

= ℓ − 3 + 3

= ℓ ✓ Step 2 Show that w(ℓ(w)) = w.

w(ℓ(w)) = w(0.5w + 3)

= 2 [ (0.5w + 3) − 3 ]

= 2(0.5w + 3 − 3)

= 2(0.5w)

= w ✓

62. y = xm/n has an inverse function except when m is even and n

is odd; Sample answer: Use a graphing calculator to graph

y = x2/3, y = x3/2, y = x2/6, and y = x3/5 to see that only

y = x2/3 does not pass the horizontal line test.

63. a. First, fi nd the inverse.

C = 5 —

9 (F − 32)

9 — 5 C = F − 32

9 — 5 C + 32 = F

The function converts temperatures from degrees Celsius

to degrees Fahrenheit.

b. The temperature at the start of the race was

F = 9 —

5 (5) + 32 = 41°F and at the end of the race was

F = 9 —

5 (−10) + 32 = 14°F.

c.

100

−100

−100

100

F = C + 3295

C = (F − 32)59

The temperature is the same at −40°.

64. a. Step 1 Find the inverse.

A = 0.2195h0.3964

A —

0.2195 = h0.3964

( A —

0.2195 )

1/0.3964

= h

Step 2 Find h when A = 1.6.

h = ( 1.6 —

0.2195 ) 1/0.3964

≈ 150.1

The height of the person is about 150.1 centimeters.

b. Step 1 Show that A(h(A)) = A.

A(h(A)) = A ( ( A —

0.2195 )

1/0.3964

) = 0.2195 [ ( A

— 0.2195

) 1/0.3964 ] 0.3964

= 0.2195 ( A —

0.2195 )

= A ✓

Step 2 Show that h(A(h)) = h.

h(A(h)) = h(0.2195h0.3964)

= ( 0.2195h0.3964

—— 0.2195

) 1/0.3964

= (h0.3964)1/0.3964

= h ✓

65. B; When you refl ect the 66. C; When you refl ect the

graph of f (x) = 3 √—

x − 4 graph of f (x) = 3 √—

x + 4

in the line y = x, you get in the line y = x, you get

the graph shown in B. the graph shown in C.

x

y4

2

−4

62−2

x

y

2

−2

−2−6

67. A; When you refl ect the graph of f (x) = √—

x + 1 − 3 in the

line y = x, you get the graph shown in A.

x

y2

−4

2−2

68. D; When you refl ect the graph of f (x) = √—

x − 1 + 3 in the

line y = x, you get the graph shown in D.

x

y

4

6

2

4 62

69. a. false; All functions of the form f (x) = xn, where n is an

even integer, fail the horizontal line test.

b. true; All functions of the form f (x) = xn, where n is an

odd integer, pass the horizontal line test.

70. Sample answer: Three points on the graph of the inverse

function are (−4, 0), (−2, 1), and (0, 2). These are on the

graph of the inverse because the roles of x and y switch and

the points (0, −4), (1, −2), and (2, 0) lie on the graph of f.

282 Algebra 2 Copyright © Big Ideas Learning, LLC. Worked-Out Solutions All rights reserved.

Chapter 5

71. Find the inverse.

y = mx + b

x = my + b

x − b = my

1 —

m (x − b) = y

So, the inverse of f is g(x) = 1 —

m (x − b), which is a linear

function. The slope is 1 —

m and the y-intercept is

−b —

m .

72. a.

x

y

4

5

2

3

1

−3

−4

−5

−2

4 5321−2−1−3−4−5

f(x) = −x

The function f (x) = −x is its own inverse because it is a

refl ection of itself in the line y = x. Find the inverse of

f (x) = −x.

y = −x

x = −y

−x = y

So, the inverse of f is g(x) = −x.

b. Sample answer:

x

y4

−4

−2

4−2−4

y = −x + 1

y = −x + 2

The functions y = −x + 1 and y = −x + 2 are their own

inverses.

c. In general, the linear functions that are their own inverses

have the form y = −x + b or y = x.


73. (−3)−3 = 1 —

(−3)3 74. 23 ⋅ 22 = 23 + 2 = 25

= − 1 —

33

75. 45

— 43

= 45 − 3 = 42 76. ( 2 — 3 )

4

= 24

— 34

77. The function is increasing when x > 1, and decreasing when

x < 1. The function is positive when x < 0 and x > 2, and

negative when 0 < x < 2.

78. The function is increasing when x < −1.15 and x > 1.15,

decreasing when −1.15 < x < 1.15. The function is positive

when −2 < x < 0 and x > 2, and negative when x < −2 and

0 < x < 2.

79. The function is increasing when −2.89 < x < 2.89,

decreasing when x < −2.89 and x > 2.89. The function is

positive when x < −5 and 0 < x < 5, and negative when

−5 < x < 0 and x > 5.

5.4 –5.6 What Did You Learn? (p. 285)

1. In the case of the minimum wind speed, the Beaufort number

will be 0. In the case of the maximum wind speed, the

Beaufort number will be 12.

2. Knowing that the result of the square root will always be

positive allows the conclusion that the equation will have no

solution when the constant is negative.

3. Calculating how long it would take Elvis to reach the ball if

he ran from point A to point C then swam to point B gives

16.46 seconds. Any other path to the ball should result in a

time shorter than this.

4. You can graph the pairs of functions and then determine

whether they are refl ections in the line y = x.

Chapter 5 Review (pp. 286–288)

1. 87/3 = (81/3)7 = 27 = 128

2. 95/2 = (91/2)5 = 35 = 243

3. (−27)−2/3 = [ (−27)1/3 ] −2 = (−3)−2 =

1 —

(−3)2 =

1 —

9

4. x5 + 17 = 35 5. 7x3 = 189

x5 = 18 x3 = 27

x ≈ 1.78 x = 3

The solution is x ≈ 1.78. The solution is x = 3.

6. (x + 8)4 = 16

x + 8 = ±2

x = −8 ± 2

The solutions are x = −10 and x = −6.

7. ( 61/5

— 62/5

) 3

= (61/5)3

— (62/5)3

= 63/5

— 66/5

= 1 —

66/5 − 3/5

= 1 —

63/5

Copyright © Big Ideas Learning, LLC. Algebra 2 283All rights reserved. Worked-Out Solutions

Chapter 5

8. 4 √—

32 ⋅ 4 √—

8 = 4 √—

32 ⋅ 8 = 4 √—

256 = 4

9. 1 —

2 − 4 √—

9 =

1 —

2 − 4 √—

9 ⋅

2 + 4 √—

9 —

2 + 4 √—

9

= 2 +

4 √—

9 —

1

= 2 + 4 √—

32

= 2 + √—

3

10. 4 5 √—

8 + 3 5 √—

8 = 7 5 √—

8

11. 2 √—

48 − √—

3 = 2 √—

16 ⋅ 3 − √—

3

= 2 √—

16 ⋅ √—

3 − √—

3

= 2 ⋅ 4 ⋅ √—

3 − √—

3

= 8 √—

3 − √—

3

= 7 √—

3

12. (52/3 ⋅ 23/2)1/2 = (52/3)1/2 ⋅ (23/2)1/2

= 52/6 ⋅ 23/4

= 51/3 ⋅ 23/4

13. 3 √—

125z9 = 3 √—

125 ⋅ 3 √—

z9 = 5z3

14. 21/4z5/4

— 6z

= 21/4z5/4 − 1

— 6

= 21/4z1/4

— 6

= (2z)1/4

— 6

15. √—

10z5 − z2 √—

40z = z2 √—

10z − 2z2 √—

10z

= (z2 − 2z2) √—

10z

= −z 2 √—

10z

16. The graph of g is a vertical stretch by a factor of 2 followed

by a refl ection in the x-axis of the graph of f.

x

y

2

1

3

4

−3

−4

−5

−6

−7

−8

−2

4 5 6 7 8 9 10321−2−1

g

f

17. The graph of g is a refl ection in the y-axis and a translation

6 units down of the graph of f.

x

y

2

1

3

4

−3

−4

−6

−7

−8

−2

4 5 6 7 8321−2−1−3−4

g

f

18. Step 1 First write a function h that represents the refl ection

of f.

h(x) = f (−x)

= 3 √—

−x


of h.

g(x) = h(x − 7)

= 3 √—

−(x − 7)

= 3 √—

−x + 7


−x + 7 .


2y2 = x − 8

y2 = 1 —

2 x − 4

y = ± √— 1 —

2 x − 4

Step 2 Graph both radical functions.

y1 = √— 1 —

2 x − 4

y2 = − √— 1 —

2 x − 4

14

−5

5

−1

y = ± x − 82

The vertex is (8, 0) and the parabola opens right.


Chapter 5


x2 + y2 = 81

y2 = 81 − x2

y = ± √—

81 − x2

Step 2 Graph both radical functions using a square viewing

window.

y1 = √—

81 − x2

y2 = − √—

81 − x2

15

−10

10

−15

y = ± 81 − x2



21. 4 3 √—

2x + 1 = 20 Check:

3 √—

2x + 1 = 5 4 3 √—

2(62) + 1 ==?

20

( 3 √—

2x + 1 ) 3 = 53 4 3 √—

125 ==?

20

2x + 1 = 125 4 ⋅ 5 ==?

20

2x = 124 20 = 20


22. √—

4x − 4 = √—

5x − 1 − 1

( √—

4x − 4 ) 2 = ( √—

5x − 1 − 1 ) 2

4x − 4 = ( √—

5x − 1 ) 2 − 2 √—

5x − 1 + 1

4x − 4 = 5x − 1 − 2 √—

5x − 1 + 1

−x − 4 = −2 √—

5x − 1

x + 4 = 2 √—

5x − 1

(x + 4)2 = ( 2 √—

5x − 1 ) 2

x2 + 8x + 16 = 4(5x − 1)

x2 + 8x + 16 = 20x − 4

x2 − 12x + 20 = 0

(x − 10)(x − 2) = 0

x − 10 = 0 or x − 2 = 0

x = 10 or x = 2

Check:

√—

4(10) − 4 ==?

√—

5(10) − 1 − 1

√—

36 ==?

√—

49 − 1

6 ==?

7 − 1

6 = 6 ✓

√—

4(2) − 4 ==?

√—

5(2) − 1 − 1

√—

4 ==?

√—

9 − 1

2 ==?

3 − 1

2 = 2 ✓


23. (6x)2/3 = 36

[ (6x)2/3 ] 3/2 = 363/2

6x = ±216

x = ±36

Check:

(6 ⋅ 36)2/3 ==?

36 [ 6 ⋅ (−36) ] 2/3 ==?

36

(216)2/3 ==?

36 (−216)2/3 ==?

36

36 = 36 ✓ 36 = 36 ✓

The solutions are x = −36 and x = 36.

24. Step 1 Solve for x.

5 √—

x + 2 > 17

5 √—

x > 15

√—

x > 3

x > 9

Step 2 Consider the radicand.

x ≥ 0

So, the solution is x > 9.


2 √—

x − 8 < 24

√—

x − 8 < 12

x − 8 < 144

x < 152


x − 8 ≥ 0

x ≥ 8

So, the solution is 8 ≤ x < 152.


7 3 √—

x − 3 ≥ 21

3 √—

x − 3 ≥ 3

x − 3 ≥ 27

x ≥ 30


x − 3 ≥ 0

x ≥ 3


27. s(d) = √—

9.8d

200 = √—

9.8d

40,000 = 9.8d

4082 ≈ d

The depth of the water is about 4082 meters.


Chapter 5

28. ( fg)(x) = f (x)g(x) = ( 2 √—

3 − x ) ( 4 3 √—

3 − x ) = 8(3 − x)5/6

The function f has the domain x ≤ 3 and g has the domain of

all real numbers. So, the domain of fg is x ≤ 3. When x = 2,

the value of the product is

( fg)(2) = 8(3 − 2)5/6 = 8(1)5/6 = 8(1) = 8.

( f — g ) (x) =

2 √—

3 − x —

4 3 √—

3 − x =

1 —

2 (3 − x)1/6

The function f has the domain x ≤ 3 and g has the domain

of all real numbers. So, the domain of f —

g is x < 3.

When x = 2, the value of the quotient is

( f — g ) (2) =

1 —

2 (3 − 2)1/6 =

1 —

2 (1) =

1 —

2 .

29. ( f + g)(x) = f (x) + g(x) = (3x2 + 1) + (x + 4)

= 3x2 + x + 5


numbers. So, the domain of f + g is of all real numbers.

When x = −5, the value of the sum is

( f + g)(−5) = 3(−5)2 + (−5) + 5 = 75.

( f − g)(x) = f(x) − g(x) = (3x2 + 1) − (x + 4) = 3x2 − x − 3


numbers. So, the domain of f − g is of all real numbers.

When x = −5, the value of the difference is

( f − g)(−5) = 3(−5)2 − (−5) − 3 = 77.

30. f (x) = − 1 — 2 x + 10

y = − 1 — 2 x + 10

x = − 1 — 2 y + 10

x − 10 = − 1 — 2 y

−2(x − 10) = y

The inverse of f is g(x) = −2(x − 10), or g(x) = −2x + 20.

x

y

15

20

5

20155−5

g

f

31. f (x) = x2 + 8, x ≥ 0

y = x2 + 8

x = y2 + 8

x − 8 = y2

± √—

x − 8 = y

Because the domain of

f is x ≥ 0, the range of the

x

y

4

8

12

16

8 12 164

g

f

inverse is y ≥ 0. So,

the inverse of f is

g(x) = √—

x − 8 .

32. f (x) = −x3 − 9

y = −x3 − 9

x = −x3 − 9

x + 9 = −y3

−x − 9 = y3

3 √—

−x − 9 = y


−x − 9 .

x

y

4

−12

4−12

f

g

33. f (x) = 3 √—

x + 5

y = 3 √—

x + 5

x = 3 √—

y + 5

x − 5 = 3 √—

y

x − 5

— 3 = √

— y

( x − 5 —

3 )

2

= y

1 —

9 (x − 5)2 = y

Because the range of f is y ≥ 5, the domain of g is x ≥ 5.

So, the inverse of f is g(x) = 1 —

9 (x − 5)2, x ≥ 5.

x

y

4

2

6

8

10

12

14

8 10 12 14 16 18642

16

18

f

g

34. Step 1 Show that f(g(x)) = x.

f(g(x)) = f ( 1 — 4 (x + 11)2 )

= 4 ( 1 — 4 (x + 11)2 − 11 ) 2

≠ x

So, the functions are not inverses.


Chapter 5


f (g(x)) = f ( − 1 — 2 x + 3 )

= −2 ( − 1 — 2 x + 3 ) + 6

= x − 6 + 6

= x ✓

Step 2 Show that g( f (x)) = x.

g( f (x)) = g(−2x + 6)

= − 1 — 2 (−2x + 6) + 3

= x − 3 + 3

= x ✓


36. Step 1 Find the inverse.

d = 1.587p

d —

1.587 = p

Step 2 Evaluate the inverse when d = 100.

p = 100

— 1.587

≈ 63

So, 100 U.S. dollars is about 63 British pounds.

Chapter 5 Test (p. 289)

1. Inequality: 5 √—

x − 3 − 2 ≤ 13

5 √—

x − 3 ≤ 15

√—

x − 3 ≤ 3

x − 3 ≤ 9

x ≤ 12


x − 3 ≥ 0

x ≥ 3

So, the solution is 3 ≤ x ≤ 12.

Equation: 5 √—

x − 3 − 2 = 13

5 √—

x − 3 = 15

√—

x − 3 = 3

x − 3 = 9

x = 12

The steps used to solve a radical inequality and to solve

a radical equation are the same, but when solving the

inequality, you must check the domain for values that make

the radicand negative.

2. The graph of g is a translation 3 units right of the graph of f.

g(x) = f (x − 3)

= √—

x − 3


by a refl ection in the x-axis of the graph of f.

g(x) = −2f(x)

= −2 3 √—

x


by a translation 2 units up the graph of f.

g(x) = 2f(x) + 2

= 2 5 √—

x + 2

5. 642/3 = (641/3)2 = 42 = 16


6. (−27)5/3 = [ (−27)1/3 ] 5 = (−3)5 = −243

The cube root of −27 is −3 and (−3)5 is −243.

7. 4 √—

48xy11z3 = 2y2 4 √—

3xy3z3

The fourth root of 16 and y8 can be simplifi ed.

8. 3 √—

256 —

3 √—

32 = 3

√—

256

— 32

= 3 √—

8 = 2

The radical expression can be simplifi ed to 3 √—

8 and the cube

root of 8 is 2.

9. Sample answer: y = √—

x − 4 is a translation of the graph of

y = √—

x 4 units right and has a domain of x ≥ 4. y = √—

x − 2

is a translation of the graph of y = √—

x 2 units down and has

a range of y ≥ −2.

10. h = 0.9(200 − a)

h —

0.9 = 200 − a

h —

0.9 − 200 = −a

− h —

0.9 + 200 = a

When h = 36: a = − 36

— 0.9

+ 200

= −40 + 200

= 160

The average score is 160.

11. Rabbit: R = 73.3(2.5)3/4

≈ 145.7 Kcal/day

Sheep: R = 73.3(50)3/4

≈ 1378.3 Kcal/day

Human: R = 73.3(70)3/4

≈ 1773.9 Kcal/day

Lion: R = 73.3(210)3/4

≈ 4043.6 Kcal/day


Chapter 5

12. ( f + g)(x) = f (x) + g(x) = 6x3/5 − x3/5 = 5x3/5

The functions f and g each have the same domain: all

real numbers. So, the domain of f + g is all real numbers.

When x = 32, the value of the sum is

( f + g)(32) = 5(32)3/5 = 5(8) = 40.

( f − g)(x) = f (x) − g(x) = 6x3/5 + x3/5 = 7x3/5

The functions f and g each have the same domain: all

real numbers. So, the domain of f − g is all real numbers.

When x = 32, the value of the difference is

( f − g)(32) = 7(32)3/5 = 7(8) = 56.

13. ( fg)(x) = f(x)g(x) = ( 1 — 2 x3/4 ) (8x) = 4x7/4


numbers. So, the domain of fg is x ≥ 0. When


( fg)(16) = 4(16)7/4 = 4(128) = 512.

( f — g ) (x) =

1 —

2 x3/4

— 8x

= 1 —

16x1/4



g consists of x > 0. When

x = 16, the value of the quotient is

( f — g ) (16) =

1 —

16(16)1/4 =

1 —

16(2) =

1 —

32 .

14. h = 1 —

64 s2

64h = s2

√—

64h = s

8 √—

h = s

The initial speed of the player is about

8 √—

3 ≈ 13.9 feet per second.

Step 1 Show that h(s(h)) = h.

h(s(h)) = h ( 8 √—

h )

= 1 —

64 ( 8 √

— h ) 2

= 1 —

64 (64h)

= h ✓

Step 2 Show that s(h(s)) = s.

s(h(s)) = s ( 1 — 64

s2 ) = 8 √—

1 —

64 s2

= 8 ⋅ 1 —

8 s

= s ✓

Chapter 5 Standards Assessment (pp. 290–291)

1. The pairs of equivalent expressions are

1. a and n √—

an because n √—

an = an/n = a.

2. a1/n and n √—

a because a1/n = n √—

a .

3. ( √—

a ) n and √—

an because ( √—

a ) n = √—

an .

2. The parent function is y = x2. The graph has been

translated 3 units left and 2 units up. So, the function is

f(x) = (x − (−3))2 + 2.

3. a. n = 2: s = 4.62 9 √—

2 ≈ 5.0 m/sec

n = 4: s = 4.62 9 √—

4 ≈ 5.4 m/sec

n = 8: s = 4.62 9 √—

8 ≈ 5.8 m/sec

b. no; the boat speed does not double when the number

of people are doubled because of the ninth root, so it

increases by a factor of 21/9.

c. n = 2: 2000

— 4.99

≈ 400.8 sec = 400.8 sec ⋅ 1 min

— 60 sec

≈ 6.7 min

n = 4: 2000

— 5.39

≈ 371.1 sec = 371.1 sec ⋅ 1 min

— 60 sec

≈ 6.2 min

n = 8: 2000

— 5.82

≈ 343.6 sec = 343.6 sec ⋅ 1 min

— 60 sec

≈ 5.7 min

4. 28 2 −6

26 8 −4 −10 −10

18

6 6 6

12 6 0

−2 8 18

The third differences are constant, so the degree of the

polynomial is 3. Work backwards to fi nd the missing values

in the table.

−2 8 18

−10 −10 −4 8

0

6 6

−6 −12

22 14 5. 18 minus what number is −4?

6. 22 minus what number is 8?

3. −10 minus what number is −6?

4. −4 minus what number is −12?

1. 0 minus what number is 6?

2. −6 minus what number is 6?

The missing values are 22 and 14.

5. 42 = 1 —

2 (x + 8)x

84 = (x + 8)x

84 = x2 + 8x

0 = x2 + 8x − 84

0 = (x − 6)(x + 14)

x − 6 = 0 or x + 14 = 0

x = 6 or x = −14

Reject the negative solution because a negative length does

not make sense. So, x = 6.


Chapter 5

6. Equation Parabola Function

y = (x + 3)2 ✓ ✓

x = 4y2 − 2 ✓

y = (x − 1)1/2 + 6 ✓

y2 = 10 − x2

Parabolas are of the form y = x2 or x = y2. There are

two equations in the table that are parabolas. Two of the

equations represent functions because they pass the Vertical

Line Test but the other two do not.

7. C; 2 √—

x + 3 − 1 < 3

2 √—

x + 3 < 4

√—

x + 3 < 2

x + 3 < 4

x < 1


x + 3 ≥ 0

x ≥ −3

So, the solution is −3 ≤ x < 1.

8. C; The graph is a horizontal shrink by a factor of 1 —

2 and a

translation 1 unit down of the parent absolute value function.

9. d = √—

2500 + h2

d2 = 2500 + h2

d2 − 2500 = h2

√—

d2 − 2500 = h

When d = 100: h = √——

1002 − 2500 = √—

7500 ≈ 87

So, the height of the balloon is about 87 feet.

10. They are not inverse functions because they are not

refl ections of each other in the line y = x.

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White Plains Middle School€¦ · 238 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved. Chapter 5 35. x 3 = 125 36. 5x = 1080 x = 5 x3 = 216