CHAPTER
8
Solutions Key
Rational and Radical Functions
8-1 VARIATION FUNCTIONS,
PAGES 569-576
ARE YOU READY? PAGE 565
1. D
2. A
3. B
4. F
1.
5. C
6.
CHECK IT OUT!
x 11y 5
_
x 4y 7
x 11 - 4y 5 - 7
x 7y -2
7
x
_
7.
( _)
3x 2 y
z
4
(3x 2y) 4
_
2.
3
5
(3 · 6)(x y · xy )
P
P1 _
_
= 2
s1
s2
18 = _
75
_
s
1.5
18s = 75(1.5)
s = 6.25
The side length s is 6.25 in.
4 6
18x y
3
-4
(2 )(x )
8x -12
8
_
L = kr
63π = k(3.5)(18)
π=k
The radius r is 1.6 m.
k
4. y = _
x
k
4=_
10
k = 40
40
y=_
x
3.
3
11. 12x 0
12(1)
12
x 12
12. 5x 2 + 6x + 6
13. -7x + 12
14. 4x 2 - 3x
15. 3a 2 = 3 · a · a
12a = 2 · 2 · 3 · a
The GCF of 3a 2 and 12a is 3a.
4
17. 16x = 2 · 2 · 2 · 2 · x · x · x · x
40x 3 = 2 · 2 · 2 · 5 · x · x · x
The GCF of 16x 4 and 40x 3 is 8x 3.
18. (x - 5)(x + 1)
19. (x - 4)(x + 6)
20. (x + 4)(x + 8)
21. (x + 3)(x + 6)
22. (x - 3) 2
23. (x - 10)(x + 2)
2
24. 5x = 45
x2 = 9
= ±3
x = ± √9
2
25. 4x - 7 = 93
4x 2 = 100
x 2 = 25
x = ± √
25 = ±5
L = πr
8π = πr(5)
1.6 = r
y
x
1250 = _
250
k =_
5. t = _
v
15
3
2
16. c d = c · c · d
cd 2 = c · d · d
The GCF of c 2d and cd 2 is cd.
9. (3x 3y)(6xy 5)
3
x
x6
-4
10. (2x )
y
z4
-2
z4
8 4
81x y
_
y2
3
8. (x )
3(-2)
x
x -6
1
_
y = kx
6.5 = 13x
0.5 = k
y = 0.5x
1 hours to build a house.
It would take them 83 _
3
b. direct
6a. inverse
0.05(400)
kT
_
_
=
7. V =
= 20
1
P
The volume is 20 L.
THINK AND DISCUSS
1. Possible answer: A direct variation equation is in the
form y = mx + b, with m = k and b = 0.
2. Possible answer: The length varies inversely as the
width, with a constant of variation of 400.
26. 2(x - 2) 2 = 32
(x - 2) 2 = 16
x - 2 = ± √
16
x=2±4
x = 6 or -2
293
Holt McDougal Algebra 2
3.
4YPEOF6ARIATION
%QUATION
$IRECT
y kx
'RAPH
14
9. y = _
x
%XAMPLE
y
32
10. y = _
x
d rt
x
y
y
x
x
*OINT
y kxz
)NVERSE
k
y_
x
I Prt
y
5
11. y = -_
x
V
I_
R
x
y
x
EXERCISES
GUIDED PRACTICE
1. indirect variation
12.
3. y = -9x
2. y = 2x
y
y
x
x
d
t=_
r
d
4.75 = _
60
285 = d
285 = 5.7
d =_
t=_
r
50
It would take the driver 5.7 h.
13. neither
14. inverse
15. direct
4. y = 12x
16.
y
x
5.
PRACTICE AND PROBLEM SOLVING
λ1 _
λ
_
= 2
v1
v2
60 = _
λ
_
1x
17. y = _
2
t1
t2
8.
V = kw
224 = k(8)(4)
7=k
The length is 6 ft.
x
x
178.25
=_
t
15
116.25t = 178.25(15)
t = 23
The time t is 23 h.
7.
y
y
d1 _
d
_
= 2
116.25
_
18. y = 6x
3
15
3(60) = 15λ
12 = λ
The wavelength λ is 12 ft.
6.
kd
P=_
t
k(500)
_
147 =
50
14.7 = k
14.7(700)
kd = _
= 343
P=_
t
30
The power P is 343 kilowatts.
19. y = -3x
V = kw
210 = 7(5)
6=
y
x
C = mtk
12 = m(50)(6)
0.04 = m
C = mtk = 0.04(30)(8) = 9.6 ≈ 10
The total cost C is about 10 cents.
294
Holt McDougal Algebra 2
d2
d1
_
=_
m1
m2
100 = _
d
_
20.
k
s=_
t
_
200 = k
31.5
6300 = k
6300
s=_
t
6300
_
b. s =
t
6300
210 = _
t
30 = t
It takes 30 s to complete one lap.
37a.
55
70
100(70) = 55d
127 ≈ d
The dosage d is 127 mg.
C1 _
C
_
= 2
w1
w2
C
25 = _
_
21.
22.
23.
3.2
12.35
25(12.35) = 3.2C
88 ≈ C
There are 88 Cal in the melon.
38. Answers will vary.
N = kap
980 = k(700)(70)
0.02 = k
N = kap = 0.02(1000)(75) = 1500
The number of bags N is 1500.
39a.
Q = kmT
20930 = k(1)(5)
4186 = k
The mass m is 0.2 kg.
4
24. y = _
5x
1150 , where I is intensity
Possible answer: I(d) = _
d2
in milliwatts per square centimeter and d is distance
from the light source in centimeters
b. True Federal Bank
y
y
x
( )
6 = 30
c. I = kPt = 0.02(3000) _
12
The interest earned in six months is $30.
40. 12; -16; 0.04
x
41. x = 5; z = 2; y = 4.4
42. approximately 9 min
43. Possible answer: 2 points are needed to determine
a line. Since direct variations always include (0, 0),
only 1 other point is needed to write the equation.
6
26. y = -_
x
( )
3
12.5 = k(2500) _
12
0.02 = k
I = 0.02Pt
Q = kmT
8372 = 4186m(10)
0.2 = m
10.5
25. y = _
x
I = kPt
44. Possible answer: If the ratios of the coordinates
in each ordered pair are the same, the variation is
direct. If the products of the coordinates in each
ordered pair are the same, the variation is inverse.
y
x
TEST PREP
45. D
47. D
k
27. d = _
w
k
3=_
20
60 = k
60 = 5
k =_
d=_
w
12
It would take 5 days.
28. inverse
48.
k
k
c=_
c=_
n
n
6000
k
250 = _
200 = _
n
24
n = 30
6000 = k
The number of students n is 30.
CHALLENGE AND EXTEND
29. neither
49.
y = kxz 2
189 = k(7)(9 )
_1 = k
3
1 (2)(6 2) = 24
y = kxz 2 = _
3
50a. r = kp
19 = k(12,281,054)
k ≈ 0.00000155
r ≈ 0.00000155p
2
30. direct
31.
46. H
kT
kT
V=_
V=_
P
P
k(320)
0.0625(330)
_
__
20 =
15 =
1
P
P = 1.375
0.0625 = k
The pressure P is 1.375 atm.
32. always
33. sometimes
34. never
35. always
36. never
295
Holt McDougal Algebra 2
b. FL: r ≈ 0.00000155(15,982,378) ≈ 25
IL: r ≈ 0.00000155(12,419,293) ≈ 19
MI: r ≈ 0.00000155(9,938,444) ≈ 15
8-2 MULTIPLYING AND DIVIDING
RATIONAL EXPRESSIONS,
PAGES 577-582
c. r = kp
32 ≈ 0.00000155p
p ≈ 21,000,000
The population p is about 21,000,000
7πz =
51. y = _
x
2
2
7π(2 )
_
12
CHECK IT OUT!
16x 11
1a. _
8x 2
16
_x 11 - 2
8
2x 9; x ≠ 0
≈7
SPIRAL REVIEW
52.
h1 _
h
_
= 2
c.
t1
t2
8
_
6
985
12 = _
_
t
4
9_
12
8 t = 985 9 _
4
6_
12
12
t = 1379
The length of the Eiffel Tower’s shadow is 1379 ft.
55.
(2x - 3)(3x + 2)
2x - 3
3
b.
1
2
-x + 3x
__
g(x)
63
-_
32
15
-_
8
3
-_
2
0
6
y
x
2x - 7x + 3
x 2 - 6x + 6 5x + 15
10(x - 4)
x+3
__
·_
(x - 4)(x - 2) 5(x + 3)
10 · _
1
_
x-2 5
2
_
x-2
-2
-1
0
1
2
h(x)
65
_
17
_
_5
64
16
4
2
5
y
5a.
2
12y 2
x ·_
_
4
3y
_
x 4y
x2
2
x 2 + x - 12
_
= -7
b.
x+4
(x + 4)(x - 3)
__
= -7
x+4
x - 3 = -7
x = -4
since x ≠ -4, no solution
x
2
x 4y
x ÷_
4a. _
4
12y 2
4x - 1
2x - 7x - 4 ÷ __
b. __
x2 - 9
8x 2 - 28x + 12
4(x - 3)(2x - 1)
(x - 4)(2x + 1) __
__
·
(x + 3)(x - 3) (2x + 1)(2x - 1)
4(x - 3)
(x - 4)(2x + 1) _
__
·
2x + 1
(x + 3)(x - 3)
4(x - 4)
_
x+3
asymptote: y = -2; vertically compressed by
1 and translated 2 units down
a factor of _
2
x
2
x+3
10x - 40 · _
_
2
56.
20
x ·_
x ·_
3a. _
15 2x x 4
7
20
x ·_
_
20 x 4
3
2x
_
3
7
2
-x(x - 3)
__
b.
10 - 2x
2a. _
x-5
-2(x - 5)
_
x-5
-2; x ≠ 5
2
2
2x - 1
0
6x - 5x - 6
(2x - 1)(x - 3)
-1
2
-x ; x ≠ 3 and x ≠ _
1
_
-2
2
6x + 7x + 2
__
2x + 1
3
2 and x ≠ _
_
; x ≠ -_
1x + 8
54. y = -_
2
x
3x 2 + x - 4
3x + 4
__
(3x + 4)(x - 1)
4 and x ≠ 1
1 ; x ≠ -_
_
3
x-1
(2x + 1)(3x + 2)
__
( )
5x + _
1
53. y = _
4
4
3x + 4
_
b.
asymptote: y = 1; translated 1 unit right and
1 unit up
2
4x - 9 = 5
_
(2x + 3)
(2x + 3)(2x - 3)
__
=5
2x + 3
2x - 3 = 5
2x = 8
x=4
THINK AND DISCUSS
1. Possible answer: Set the denominator of the
unsimplified expression equal to 0, and solve.
2. Possible answer: Solving a rational equation may
produce extraneous solutions.
296
Holt McDougal Algebra 2
3.
&RACTIONS
2ATIONAL%XPRESSIONS
3IMPLIFYING
?
?
?
2
2
x + 2x + 1
x -1
__
÷_
y
x
u ?
?
?
u ?
?
$IVIDING
14.
x y x y x
?
? ?
xy xy y y -ULTIPLYING
2
2
x - 25
x - 3x - 10
__
÷ __
x
13.
x y y y x u ?
x u ?
?
?
?
y x
y x
y
x ?
x u ?
x µ ?
?
?
y
y y x
y ?
?
µ?
u?
?
y
EXERCISES
GUIDED PRACTICE
1. Possible answer: A rational expression is the
quotient of 2 polynomials.
2.
4.
6.
4x 6
_
2x - 6
6
2(2x )
_
2(x - 3)
6
2x ; x ≠ 3
_
x-3
x+4
__
3x 2 + 11x - 4
x+4
__
(x + 4)(3x - 1)
1 ;
_
3x - 1
1
x ≠ -4 and x ≠ _
3
6x 2 + 7x - 3
__
-3x 2 + x
(2x + 3)(3x - 1)
__
-x(3x - 1)
2x + 3
_
;
-x
1
x ≠ 0 and x ≠ _
3
4x - 6
x-2 ·_
8. _
2x - 3 x 2 - 4
2(2x - 3)
x - 2 · __
_
2x - 3 (x - 2)(x + 2)
2
_
x+2
6x + 13x - 5
__
6x - 23x + 7
(2x - 7)(3x - 1)
2x + 5
7
1 and x ≠ _
_
;x≠_
7.
2
-x - 4
_
x 2 - x - 20
-(x + 4)
__
(x -5)(x + 4)
-1 ;
_
x-5
x ≠ 5 and x ≠ -4
x2 + 1
2
6x(x + 1)
_
2x - 6
x-2 ·_
9. _
5 4
x y
1
11. _ ÷ _
5 4
x 3y
x y
_
· x 3y
3xy
7 4
x
y
_
3
2x - 3
(x + 5)(2x - 3)
__
= 10
2x - 3
x + 5 = 10
x=5
2
x -4 =1
_
17.
x-2
(x + 2)(x - 2)
__
=1
x-2
x+2=1
x = -1
4x - 8
18. _
x 2 - 2x
4(x - 2)
_
x(x - 2)
_4 ; x ≠ 0 and x ≠ 2
x
20.
x-3 x+5
2(x - 3)
x - 2x · _
_
x-3
x+5
2(x - 2)
_
x+5
3xy
4x + 3
2
2x + 7x - 15
__
= 10
PRACTICE AND PROBLEM SOLVING
3
6x + 6x
_
x 2 - 4x + 4 x 2 + 6x + 8
(x
+ 4)(x - 4) __
x-2
__
·
(x + 2)(x + 4)
(x - 2) 2
x-4
__
(x + 2)(x - 2)
x+3
x+3
12. _ ÷ _
3
x2 + 1
6x; defined for all real
values of x
2
x - 16 · _
x-2
10. _
16.
4x + 3
4x - 3 = -6
3
x = -_
4
3
since x ≠ -_, no solution
4
(2x + 5)(3x - 1)
__
5.
2
16x - 9 = -6
_
(4x - 3)(4x + 3)
__
= -6
2
2x - 7
x 2 - 3x - 18
x 2 - 7x + 6
2
(x + 1)
(x - 6)(x - 1)
__
· __
(x - 6)(x + 3) (x + 1)(x - 1)
x+1
_
x+3
15.
2
3.
2x 2 + 5x - 12
x 2 + 9x + 20
(x
+
5)(x
5)
(x
+ 4)(x + 5)
__ · __
(x + 4)(2x - 3) (x - 5)(x + 2)
2
(x + 5)
__
(2x - 3)(x + 2)
2
x - 36
__
2
x - 12x + 36
(x + 6)(x - 6)
__
(x - 6) 2
x+6
_
;x≠6
x-6
19.
8x - 4
__
21.
3x + 18
__
2x 2 + 9x - 5
4(2x - 1)
__
(x + 5)(2x - 1)
4 ; x ≠ -5 and x ≠ _
1
_
x+5
2
24 - 2x - x 2
3(x + 6)
__
-(x + 6)(x - 4)
-3 ; x ≠ -6 and x ≠ 4
_
x-4
2
4x + 20
-2x - 9x
23. _
22. _
2
-5 - x
4x - 81
4(x + 5)
_
-x(2x + 9)
__
-(x + 5)
(2x + 9)(2x - 9)
-4;
x ≠ -5
9
9
-x ; x ≠ -_
_
and x ≠ _
2
2
2x - 9
24.
x-1
x 2 - 2x + 1
x+3
x
1
_
_
·
(x - 1) 2 x + 3
1
_
x-1
2
3y 5
x y _
_
· x ·_
4xy
x2 ·
_
6 x4
5
3y
_
24 x 4
y5
_
x-4 ·_
2x - 1
25. _
x-3 x+4
(x - 4)(2x - 1)
__
(x - 3)(x + 4)
8x 2
297
Holt McDougal Algebra 2
2
4
3
4x + 4 + xy + y
5x 41. (y + 4) ÷ __
14x · _
x ÷_
40. _
xy
3
5
3
6y
12y
3
__
5
(y + 4) ·
6
12y
7x · _
_
4(x + 1) + y(x + 1)
3
3y 4 5x 2
(y + 4) · __
4
(4 + y)(x + 1)
28
x
y
_
3
_
5
x+1
2
3x + 10x + 8
x - 2x - 8 · __
26. _
9x 2 - 16
x 2 - 16
(3x + 4)(x + 2)
(x
4)(x
+
2)
__ · __
(3x + 4)(3x - 4) (x + 4)(x - 4)
2
(x + 2)
__
(3x - 4)(x + 4)
2
27.
2
4x - 20x + 25 _
__
· 3x - 12
28.
2
2
x + 4x
4x + 15x + 9 _
__
÷
x - 5 · (x - 5)
42. (x + 1) ÷ _
x-2
x - 2 · (x - 5)
(x + 1) · _
x-5
(x + 1)(x - 2)
x2 - x - 2
2x - 5
x 2 - 4x
2
3(x - 4)
(2x - 5) _
_
·
2x - 5
x(x - 4)
3(2x - 5)
_
x
s h =_
h;
43a. square prism: _
1
s2
2
πr h = _
h
cylinder: _
2
1
πr
2
2x + 1
8x 2 + 10x + 3
(4x
+
3)(x
+
3)
2x + 1
__ · _
(4x + 3)(2x + 1) x(x + 4)
x+3
_
x(x + 4)
x - 3x - 10 30.
x - 4x - 5 ÷ __
29. _
x 2 - 3x + 2
x2 - 4
(x + 2)(x - 2)
(x - 5)(x + 1) __
__
·
(x - 1)(x - 2) (x - 5)(x + 2)
x+1
_
x-1
2
2
x+2 _
1
_
÷
x-4
3x - 12
x+2
_
· 3(x - 4)
x-4
3(x + 2)
3x + 6
c.
2
x + 2x - 15
x - 2x - 3 ÷ __
31. _
x2 - x - 2
x2 + x - 6
(x + 3)(x - 2)
(x + 1)(x - 3) __
__
·
(x +1)(x - 2) (x + 5)(x - 3)
x+3
_
x+5
2
32.
2
3x + 10x + 8
__
= -2
-x - 2
34.
2
x + 3x - 28
__
= -11
(x + 7)(x - 4)
(x + 7)(x - 4)
__
= -11
(x + 7)(x - 4)
1 = -11
no solution
3
2x · _
x
36. _
3 6x - 8
3
2x · _
x
_
3 2(3x - 4)
4
x
_
3(3x - 4)
38.
x-3
(x + 3)(x - 3)
__
=5
x-3
x+3=5
x=2
-(x + 2)
-(3x + 4) = -2
3x + 4 = 2
2
x = -_
3
35.
(2r)(2h)
rh
4rh
rh
45. Student A; the student didn’t leave a 1 in the
numerator.
πr 2 ;
_
π(5r) 2
2
2
πr = _
πr
1
_
=_
2
2
25
π(5r)
π(25r )
2x + 1
4x - 3x · _
37. _
x
2
4x - 1
x(4x - 3)
2x + 1
__
·_
x
(2x + 1)(2x - 1)
4x - 3
_
2x - 1
x
1
_
÷_
2(2r) + 2(2h) _
2r + 4h _
4r + 8h _
2r + 4h
__
÷
=
·
2r + 4h
rh
=_ ·_
2rh
2r + 4h
1
=_
2
1.
The ratio would be reduced by a factor of _
2
1 at 2
v 0t + _
d
2
1 at
_
_
= v0 + _
44a. v = =
t
t
2
1
1
_
_
b. v = v 0 + at = 264 + (10)(3) = 279
2
2
The driver’s average speed was 279 ft/s.
x2 - 9 = 5
_
33.
(3x + 4)(x + 2)
__
= -2
2
s(2s + 4h)
2s + 4h
2s + 4sh
b. square prism: _ = _ = _;
2
s(sh)
sh
s h
2
πr(2r
+
2h)
2r + 2h
2πr + 2πrh
cylinder: __ = _ = _
πr(rh)
rh
πr 2h
46. No; it is true for all real numbers except x = 2. At
this value, the rational expression is undefined.
TEST PREP
47. D;
x2 + x - 2 = 0
(x - 1)(x + 2) = 0
x - 1 = 0 or x + 2 = 0
x = 1 or -2
2
48. F;
x 2 + 7x + 10
__
49. A;
2
x + 13x + 36
x - 4x
__
÷ __
2
2
x+9
x - 6x
x - 8x + 12
(x
+
4)(x + 9)
__
(x - 2)(x - 6)
(x + 2)(x + 5) __
__
·
x
+9
x(x - 6)
x(x + 2)(x - 2)
x+4
x+5
_
2
y
2x 2 · _
39. 2xy · _
y
10x
14
2x
25x - 49
y2
2(5x - 7)
1
__
·_
4x 3 · _
x
2x
(5x + 7)(5x - 7)
2
_
2x 2y 2
x(5x + 7)
2
3
x2
298
Holt McDougal Algebra 2
1x
58. y = _
2
CHALLENGE AND EXTEND
8x - 1 · __
x -4
50. _
x+2
2x 2 - 5x + 2
3
2
(2x - 1)(x - 2)
(
2
)
CHECK IT OUT!
-1
x -9
x - 16 ÷ _
52. _
x-3
x+4
(x + 4)(x - 4) _
x+4
__
÷
x-3
x2 - 9
(x + 3)(x - 3)
(x + 4)(x - 4) __
__
·
x-3
x+4
(x - 4)(x + 3)
2
8-3 ADDING AND SUBTRACTING
RATIONAL EXPRESSIONS,
PAGES 583-590
2
2(x + 5x + 25)
__
1a.
x 3 - 2x 2 + x - 2
3 2
2
x (x - 4) - (x - 4)
__
x2 - 1
x 2 - 2x + 1
(x 2 + 1)(x - 2) 3x(x 2 + x + 1) (x - 1) 2
2
(x - 1)(x + x + 1)(x + 2)(x - 2) __
(x + 1)(x - 1)
___
·
2
(x + 1)(x - 2)
3x(x 2 + x + 1)
b.
3a.
x2 + 1
SPIRAL REVIEW
x
57. y ≈ 56,800(1.39) ;
y ≈ 56,800(1.39) 8 ≈ 800,000
The number of births is about 800,000.
2
2
x - 4 = (x + 2)(x - 2)
x + 5x + 6 = (x + 2)(x + 3)
The LCM is (x + 2)(x - 2)(x + 3).
2
3x - 2
3x + _
_
2x - 2
3x - 3
3x
3x - 2
_
+_
()
(x - 1) 2
2(x + 1)(x + 2)
__
3 5
3 3
2 4
8x y - 32x y + 56x y
2
b.
2(x - 1)
3(x - 1)
3x
3x - 2 _
2
_
_3 + _
2(x - 1) 3
3(x - 1) 2
9x + 2(3x - 2)
__
6(x - 1)
9x + 6x - 4
__
6(x - 1)
15x - 4 ; x ≠ 1
_
6(x - 1)
6x
·_
56. 8x 2y 3(xy 2 - 4x + 7y)
x -3
2x - 3x - 2
3x - 5 - __
b. _
3x - 1
3x - 1
2
2
3x - 5 - (2x - 3x - 2)
___
3x - 1
x 2 + 3x - 3
1
_
;x≠_
3x - 1
3
3 7
2
3
7
2a. 4x y = 2 · x · y
5 4
5
4
3x y = 3 · x · y
The LCM is 12x 5y 7.
(x 3 - 1)(x 2 - 4) __
(x + 1)(x - 1) _
6x
__
·
·
2
55. (x + 5)(3x - 7x - 1)
3
2
3x - 7x - x + 15x 2 - 35x - 5
3x 3 + 8x 2 - 36x - 5
x -3
x2 - 3
6x
x -1
· __
·_
3
2
x (x - 2) + (x - 2)
3x + 3x + 3x (x - 1) 2
2 4
2
6x (x ) - 6x (2)
6
2
6x - 12x
2
x2 - 3
2
4
2
9x + 4
_
3
; x ≠ ± √
2
54. 6x (x - 2)
6x + 5 _
_
+ 3x - 1
6x + 5 + 3x - 1
__
5
3
2
3
2
x - 4x - x + 4
3x + 3x + 3x
6x
53. __ ÷ __ · _
2
3
x 2 - 5x + 25
x
x
x - 125
2x - 50 · __
51. _
x 3 + 125 x 2 - 10x + 25
2
(x - 5)(x + 5x + 25)
2(x + 5)(x - 5)
__
· __
(x - 5) 2
(x + 5)(x 2 - 5x + 25)
2
y
2
(2x - 1)(4x + 2x + 1) __
(x + 2)(x - 2)
__
·
x+2
2
4x + 2x + 1
59. y = -4x
y
()
2x + 6
x +_
_
x+3
x +
_
x 2 + 6x + 9
2(x + 3)
_
x+3
(x + 3) 2
2
x +_
_
x+3
x+3
x+2
_
; x ≠ -3
x+3
3x - 2 - _
2
4a. _
2x + 5
5x - 2
2x + 5
3x - 2 _
5x - 2 - _
2
_
_
2x + 5 5x - 2
5x - 2 2x + 5
(3x - 2)(5x - 2) - 2(2x + 5)
___
(2x + 5)(5x - 2)
2
16x + 4 - 4x - 10
15
x
___
(2x + 5)(5x - 2)
2
5
15x - 20x - 6 ; x ≠ -_
2
__
and x ≠ _
2
5
(2x + 5)(5x - 2)
(
299
)
(
)
Holt McDougal Algebra 2
4b.
2
2x + 64 _
_
- x-4
x+8
x 2 - 64
2
2x + 64
_
2. Possible answer:
(
)
x-8
x-4 _
-_
x+8 x-8
x 2 - 64
2
2x + 64 - (x - 4)(x - 8)
___
2ATIONAL%XPRESSIONS
x 2 - 64
2
2x + 64 - x + 12x - 32
___
!DDINGLIKEDENOMINATORS
2
x ?
x x ?
?
x x x 2
x - 64
2
x + 12x + 32
__
3UBTRACTINGUNLIKEDENOMINATORS
x-1
x+1
_
x
÷_
x
1
x -1
x+1
x-1
__
·_
x
(x + 1)(x - 1)
2
_1
x-1
3x - 3
3(x - 1)
20 · _
_
x-1
6
10
x x ?
x ?
?
Ú
x ?
?
x
x x x x x
20
_
x-1
b. _
6
_
3x - 3
20 ÷ _
6
_
x
xxx ÚÚ
x x
x x ??
xx x x ??
xx EXERCISES
GUIDED PRACTICE
1
_1 + _
x
2x
_
c.
x+4
_
1. Possible answer: A complex fraction has 1 or more
fractions in its numerator, its denominator, or both.
2. Possible answer:
x-2
1 2x(x - 2)
_1 2x(x - 2) + _
x
2x
___
x+4
_
2x(x - 2)
x-2
2(x - 2) + (x - 2)
__
2x(x + 4)
3(x - 2)
_
2x(x + 4)
(
)
( )
3x + 4
2x - 3 + _
_
4x - 1
4x - 1
2x - 3 + 3x + 4
__
4x - 1
5x + 1
1
_
;x≠_
4x - 1
4
4x + 3
4x - 3 - _
4. _
2x - 5
2x - 5
4x - 3 - (4x + 3)
__
2x - 5
-6 ; x ≠ _
5
_
2
2x - 5
2d
_
d
d +_
_
40
x xx ?
x x+1
_
2
x
-1
_
5a.
x
_
6.
x
????
x x µ ?
?
?
????
x x x x
?
ˁ?
x x 2 - 64
(x
+
8)(x + 4)
__
(x + 8)(x - 8)
x+4
_
; x ≠ -8
x+8
()
3IMPLIFYINGACOMPLEXFRACTION
45
2d(360)
__
d (360)
d (360) + _
_
40
45
720d
_
9d + 8d
720d ≈ 42.4
_
6.
5x + 3
3x - 4 - _
3. _
4x + 5
4x + 5
3x - 4 - (5x + 3)
__
4x + 5
5
-2x - 7 ; x ≠ -_
_
4
4x + 5
2 3
2
2
3
5. 4x y = 2 · x · y
4
4
4
16x y = 2 · x · y
The LCM is 16x 4y 3.
2
x - 25 = (x + 5)(x - 5)
x + 10x + 25 = (x + 5) 2
The LCM is (x + 5) 2(x - 5).
2
3x - 2 + _
2x - 3
7. _
x+6
2x - 1
x+6
3x - 2 _
2x - 3 _
2x - 1 + _
_
x + 6 2x - 1
2x - 1 x + 6
(3x - 2)(2x - 1) + (2x - 3)(x + 6)
___
(x + 6)(2x - 1)
2
2
6x - 7x + 2 + 2x + 9x - 18
___
(x + 6)(2x - 1)
2
8x + 2x - 16
__
(x + 6)(2x - 1)
17d
The average speed is 42.4 mi/h.
(
THINK AND DISCUSS
1. Possible answer: Factor each denominator. Find the
product of the different factors. If the same factor is
in both denominators, use the highest power of that
factor.
)
(
)
2
2(x + x - 8)
1
__
; x ≠ -6 and x ≠ _
(x + 6)(2x - 1)
300
2
Holt McDougal Algebra 2
3x - 4 + _
4x - 5 + _
3x - 1
2x - 1
9. _
8. _
12x + 4
3x + 1
x+3
x2 - 9
3x - 1 _
4x - 5 + _
4
_
x-3
3x - 4 + _
2x - 1 _
_
3x + 1 4
2
4(3x + 1)
x
+
3
x
-3
x -9
4x - 5 + 4(3x - 1)
__
3x - 4 + (2x - 1)(x - 3)
___
4(3x + 1)
(x + 3)(x - 3)
4x - 5 + 12x - 4
__
2
3x - 4 + 2x - 7x + 3
__
4(3x + 1)
(x+ 3)(x - 3)
16x - 9 ; x ≠ -_
1
_
2
4x - 1 ; x ≠ ±3
2
x
__
3
4(3x + 1)
(x + 3)(x - 3)
()
(
3x - 5 - _
2x - 5
10. _
2x - 5
3x + 1
3x + 1
3x - 5 _
2x - 5 _
2x - 5
_
-_
2x - 5 3x + 1
3x + 1 2x - 5
(3x - 5)(3x + 1) - (2x - 5)(2x - 5)
___
(2x - 5)(3x + 1)
2
2
9x - 12x - 5 - 4x + 20x - 25
___
(2x - 5)(3x + 1)
2
5x + 8x - 30
5
1 and x ≠ _
__
; x ≠ -_
3
2
(2x - 5)(3x + 1)
(
11.
)
(
x-4
)
(
x -4
4x - 3
2x - 3 ÷ _
_
x-2
x2 - 4
(x + 2)(x - 2)
2x - 3 · __
_
x-2
4x - 3
(2x - 3)(x + 2)
__
4x - 3
)
2x 2
3(x
+
2)
_
6.20
7.75
2d(31)
__
d (31) + _
d (31)
_
6.20
7.75
62d
_
5d + 4d
62d ≈ 6.9
_
9d
The average speed is
about 6.9 ft/s.
2
x-5 -_
3x - 5
18. _
3x + 4
3x + 4
x - 5 - (3x - 5)
__
3x + 4
4
-2x ; x ≠ -_
_
3
3x + 4
2 3
2
2
3
20. 12x y = 2 · 3 · x · y
3 2
3
2
14x y = 2 · 7 · x · y
The LCM is 84x 3y 3.
3x - 2 + _
2x
22. _
x+2
4x - 1
x+2
3x - 2 _
4x - 1 + _
2x _
_
x + 2 4x - 1
4x - 1 x + 2
(3x - 2)(4x - 1) + 2x(x + 2)
___
(x + 2)(4x - 1)
2
2
12x - 11x + 2 + 2x + 4x
___
(x + 2)(4x - 1)
2
14x - 7x + 2
1
__
; x ≠ -2 and x ≠ _
4
(x + 2)(4x - 1)
(
)
x+3 x+1
(x + 1)(x + 3)
x + 2 - (x + 1)(x + 1)
__
(x + 3)(x + 1)
2
x + 2 - x - 2x - 1
__
(x + 3)(x + 1)
2
-x - x + 1
__
; x ≠ -3 and x ≠ -1
(x + 3)(x + 1)
2
(
x+2
2(x+
2)
+ (x+ 2)
__
2d
__
d +_
d
_
21. 16x 2 - 25 = (4x + 5)(4x - 5)
4x 2 - x - 5 = (4x - 5)(x + 1)
The LCM is (4x - 5)(4x + 5)(x + 1).
x+1 x+1
x+2
__
-_ _
2x - 3
_
_
13. x - 2
4x - 3
_
x+2
(_2x )x(x + 2) + (_1x )x(x + 2)
___
2x x( x+ 2)
_
x -3 -_
2x - 5
19. _
2x + 7
2x + 7
2
3
(2x
- 5)
x
__
2x + 7
2
x - 2x + 2
7
_
; x ≠ -_
2
2x + 7
)
(
16.
2x - 3 + _
2x - 3
17. _
4x - 7
4x - 7
2x - 3 + 2x - 3
__
4x - 7
4x - 6
_
4x - 7
2(2x - 3)
7
_
;x≠_
4
4x - 7
x+4
2x + 8
3 _
_
-_
x-4 x+4
x 2 - 16
2x + 8 - 3(x + 4)
__
(x - 4)(x + 4)
2x + 8 - 3x - 12
__
(x - 4)(x + 4)
-(x + 4)
__
(x - 4)(x + 4)
-1 ; x ≠ ±4
_
x-4
x+1
x+2
12. _ - _
2
x+3
x + 4x + 3
_2 + _1
x
x
_
2x
_
2x 2
2x + 8
3
_
-_
x 2 - 16
)
15.
3x - 7
_
4x + 5
_
14.
6x - 1
_
5x - 6
6x - 1
3x - 7 ÷ _
_
4x + 5
5x - 6
3x - 7 · _
5x - 6
_
4x + 5
6x - 1
(3x - 7)(5x - 6)
__
(4x + 5)(6x - 1)
)
8x
2x - 7 + _
23. _
x-2
3x - 6
3 +_
8x
2x - 7 _
_
x-2 3
3x - 6
3(2x - 7) + 8x
__
3(x - 2)
6x - 21 + 8x
__
3(x - 2)
14x - 21
_
3(x - 2)
7(2x - 3)
_
;x≠2
3(x - 2)
()
301
(
)
5x + _
7
24. _
2
x
+
1
4x
5 +_
7
_
4x
x+1
x+1
5 _
7 _
4x
_
+_
4x x + 1
x + 1 4x
5(x + 1) + 28x
__
4x(x + 1)
33x + 5
_
;
4x(x + 1)
x ≠ -1 and x ≠ 0
(
)
( )
Holt McDougal Algebra 2
4x - 3 - _
2x - 3
25. _
x-3
x2 - 9
x+3
2x - 3 _
4x - 3
__
-_
x-3 x+3
(x + 3)(x - 3)
(4x - 3) - (2x - 3)(x+ 3)
___
(x- 3)(x+ 3)
2
4x - 3 - 2x - 3x + 9
__
(x - 3)(x + 3)
2
-2x + x + 6
__
(x - 3)(x + 3)
-(2x + 3)(x - 2)
__
; x ≠ ±3
(x - 3)(x + 3)
2x + 1
x
-_
26. _
2x + 3
2x - 3
2x + 1 _
2x + 3
2x - 3 - _
x
_
_
2x + 3 2x - 3
2x - 3 2x + 3
x(2x - 3) - (2x + 1)(2x + 3)
___
(2x - 3)(2x + 3)
2
2
2x - 3x - 4x - 8x - 3
___
(2x - 3)(2x + 3)
2
3
-2x - 11x - 3 ; x ≠ ±_
__
2
(2x - 3)(2x + 3)
(
(
)
(
)
31.
1.5
1.5
950
The average rate is 0.6°C/min.
)
)
2x - 5
_
x2 - 9
28. _
3x - 1
_
x+3
2x - 5 ÷ _
3x - 1
_
x+3
x+3
2x - 5
__
·_
(x + 3)(x - 3)
2x - 5
__
3x - 1
(x - 3)(3x - 1)
3x - 2
_
x2 - 4
__
29.
5x + 1
_
0.4
600
_
200 + 750
600 ≈ 0.6
_
x-4
x 2 - 6x + 8
x - 2 - __
2
1 _
_
x-4 x-2
(x - 4)(x - 2)
x-2-2
__
(x - 4)(x - 2)
x-4
__
(x - 4)(x - 2)
1 ; x ≠ 2 and x ≠ 4
_
x-2
x2 - 9
0.4
100(6)
__
50 (6) + _
50 (6)
_
2
1 -_
27. _
(
100
_
50 + _
50
_
x
_
x+1
30. _
x
x+_
3
2
x +x-6
x ÷_
4x
_
5x + 1
3x - 2 ÷ _
_
x
+
1
3
x2 - 4
x2 + x - 6
3
x ·_
_
(x + 3)(x - 2)
x + 1 4x
3x - 2
__
· __
3
_
5x + 1
(x + 2)(x - 2)
4(x
+ 1)
(3x
2)(x
+
3)
__
5d
3d + _
32a. _
200
185
8d
b. _
3d + _
5d
_
185
200
8d(7400)
__
3d (7400) + _
5d (7400)
_
185
200
59200d
__
120d + 185d
59200d ≈ 194
_
305d
The average speed is 194 mi/h.
33.
x
2 +_
_
x+4
(
x-3
x-3
2 _
_
)
(
)
(
)
x+4
x _
+_
x+4 x-3
x-3 x+4
2(x - 3) + x(x + 4)
__
(x + 4)(x - 3)
2
2x
- 6 + x + 4x
__
(x + 4)(x - 3)
2
+ 6x - 6
x
__
; x ≠ -4 and x ≠ 3
(x + 4)(x - 3)
x+4
2x
+_
34. _
x+6
x 2 - 36
x+4 x-6
2x
__
+_ _
x+6 x-6
(x + 6)(x - 6)
2x + (x + 4)(x - 6)
__
(x+ 6)(x - 6)
2
2x + x - 2x - 24
__
( x + 6)(x - 6)
2
x - 24 ; x ≠ ±6
__
(x + 6)(x - 6)
35.
3
2
_
+ __
x 2 - x - 20
x 2 + 7x + 12
x
+3
3
x-5
2
__
_
_
+ __
(x - 5)(x + 4) x + 3
(x + 3)(x + 4) x - 5
2(x + 3) + 3(x - 5)
__
(x - 5)(x + 4)(x + 3)
2x + 6 + 3x - 15
__
(x - 5)(x + 4)(x + 3)
5x - 9
__
; x ≠ -4, x ≠ -3 and x ≠ 5
(x - 5)(x + 4)(x + 3)
(
(5x + 1)(x + 2)
302
)
(
)
Holt McDougal Algebra 2
36.
2
7x
x
_
+_
x-5
x - 5x
2
x
7x
_
+_
x-5
x(x - 5)
2
x
7 +_
_
x-5
x-5
2
x +7
_
;
x-5
x ≠ 0 and x ≠ 5
2
37.
(
(
)
x-1 x-2
x-2 x-1
2x(x - 2) - 9(x - 1)
__
(x - 1)(x - 2)
2
2x - 4x - 9x + 9
__
(x - 2)(x - 1)
2
2x - 13x + 9
__
;
(x - 2)(x - 1)
x ≠ 1 and x ≠ 2
2x + 3
x
38. _ - _
3x + 4
9
2x - _
_
x-1
x-2
9 _
x-2 -_
x-1
2x _
_
9x + 12
x
-_
3x + 4 3
3(3x + 4)
3(2x + 3) - x
__
3(3x + 4)
6x + 9 - x
_
3(3x + 4)
5x + 9
4
_
; x ≠ -_
3
3(3x + 4)
()
2x + 3 _
3
_
39.
3x + 4
(
2x - 3
(
)
c. The number of meters that a student can clean in
15 m = 1.5 m/min
one minute is _
10 min
1600
1600
1600
1600 - _
_
_
_
48 - _
60 = __
48
60 = _
6.7 ≈ 4.4
_
1.5
1.5
1.5
1.5
The junior class will finish about 4.4 min sooner.
4
_
x+2
_
43.
x+2
_
6
x+2
4 ÷_
_
x+2
6
6
4 ·_
_
x+2 x+2
24
_
)
( )
(
(
b.
)
)
(3x - 4)(5x + 3)
2
)
( )
6x(x- 1)
7(x3)
_
6x(x - 1)
x-4
x 2 + 4x - 32
x+8
x-5 _
6
__
-_
x-4 x+8
(x - 4)(x + 8)
6 - (x - 5)(x + 8)
__
(x - 4)(x + 8)
2
6 - x - 3x + 40
__
(x - 4)(x + 8)
2
-x - 3x + 46
__
; x ≠ -8 and x ≠ 4
(x - 4)(x + 8)
x+7
10x
41. __ - _
2
2
x + 13x + 42
x + 8x + 7
x+1
x+6
x+7
10x
__
_
_
- __
(x + 7)(x + 6) x + 1
(x + 1)(x + 7) x + 6
(x + 7)(x + 1) - 10x(x + 6)
___
(x + 1)(x + 7)(x + 6)
2
2
x + 8x + 7 - 10x - 60x
___
(x + 1)(x + 7)(x + 6)
2
-9x - 52x + 7
__
; x ≠ -7, x ≠ -6 and x ≠ -1
(x + 1)(x + 7)(x + 6)
(
2
_
3x
-4
_
44.
5x + 3
2
_
÷ (5x + 3)
3x - 4
1
2
_
·_
3x - 4 5x + 3
2
__
__
x-5
6
__
-_
(
s(s + 12)
48(48 + 12)
Each senior has to clean about 6.7 m more.
1 +_
2
_
2x
3x
2
45. _
46a. _
x-1
_
_1 + _1
a
x-3
b
2(ab)
1 6x(x - 3) + _
2 6x(x - 3)
_
__
2x
3x
_1 (ab) + _1 (ab)
___
a
b
x - 1 6x(x - 3)
_
2ab
_
x-3
b+a
3(x - 3) + 4( x- 3)
3x + 4
2
_
-_
3x + 4 2x - 3
2x - 3 3x + 4
2
4x (2x - 3) - 2(3x + 4)
__
(3x + 4)(2x - 3)
3
2
8x - 12x - 6x - 8
__
(3x + 4)(2x - 3)
3
2
2(4
x - 6x - 3x - 4)
3
4 and x ≠ _
__
; x ≠ -_
3
2
(3x + 4)(2x - 3)
40.
)
19,200
19,200
_
= _ ≈ 6.7
(x + 2)
2
4x
2
_
-_
2
2x - 3
4x
_
_
b.
)
Rooms with a Width of 30 ft
Length-to-Width
Ratio
Length (ft)
Height (ft)
2:1
60
40
3:2
45
36
4:3
40
34.3
5:3
50
37.5
√2
:1
42.4
35.1
c. Both the width and the height will double.
-3 or
x +_
47. Possible answer: _
x+2
x+2
2
x - 6x + 8 _
_
+ 1
2
x+2
x -4
48. Possible answer: Using the LCD reduces the need
to simplify the sum of the rational expressions.
1600
1600 - _
42a. _
s
s + 12
s + 12
1600 _
1600 _
s
_
-_
s s + 12
s + 12 s
1600(s + 12) - 1600s
__
s(s + 12)
19,200
_
s(s + 12)
(
)
()
303
Holt McDougal Algebra 2
TEST PREP
49. D;
3 +_
5
_
3x
9x
3 +_
5
3 _
_
3x 3
9x
9+5
_
9x
14
_
9x
(
50. H;
8
5 -_
_
x+2
()
(
x+4
x+4
5 _
_
-1
-1
56. (x - y) - (x + y)
x
+
y
x-y
1 _
1 _ -_
_
x-y x+y
x+y x-y
(x + y) - (x - y)
__
(x - y)(x + y)
2y
__
(x - y)(x + y)
x+4
5 +_
57. _
x+5
x2 - x
x(x - 1)
x+4 x+5
5 _
_
+_ _
x + 5 x(x - 1)
x(x - 1) x + 5
5x(x - 1) + (x + 4)(x + 5)
___
x(x - 1)(x + 5)
2
2
5x - 5x + x + 9x + 20
___
(
)
x+2
8 _
-_
x+2 x+4
x+4 x+2
5(x + 4) - 8(x + 2)
__
(x + 4)(x + 2)
-3x + 4
__
(x + 4)(x + 2)
51. A;
8
_
7x
_
-4
_
x+1
)
54.
(
)
)
2
= 6x + 4x + 20
so,
SPIRAL REVIEW
58.
2809.01d
(
2
-x
_
y2 - x2
2
-(-2)
__
m - nm
59. _
n 2 + 10
2
(-4) - (-4)(0)
__
(3) 2 - (-2) 2
-4
_
9-4
4
-_
5
(0) 2 + 10
16 - 0
_
0 + 10
_8
5
x
-y
y
x
61. The asymptote is x = -4. The transformation is
a translation 4 units left.
y
x
x -1(xy) + y -1(xy)
__
-1
-1
x (xy) - y
y+x
_
y-x
2
60. The asymptote is x = 1. The transformation is
a vertical stretch by a factor of 2 and a translation
1 unit right.
x -1 + y -1
_
-1
)
x 3 + 4x 2 - 5x
796.16d + 908.8d + 1104.05d
x+2
x-2
x2 - 4
x+2
6x _
x - 2 + __
4
x-1 _
_
-_
x+2 x-2
x-2 x+2
(x + 2)(x - 2)
(x - 1)(x - 2) + 4 - 6x(x + 2)
___
(x - 2)(x + 2)
2
2
x - 3x + 2 + 4 - 6x - 12x
___
(x - 2)(x + 2)
2
-5x - 15x + 6
__
(x + 2)(x - 2)
(
x(x 2 + 4x - 5)
6x + 4x + 20
__
84791.04d ≈ 30.2
_
6x
x-1 +_
4
53. _
+ _
)
)
2
84791.04d
___
CHALLENGE AND EXTEND
(
(
x+1
8 ·_
_
7x
-4
-2(x + 1)
_
7x
52. H;
3d
__
d +_
d +_
d
_
35.5
31.1
25.6
3d(28263.68)
____
d
d
d
___
___
(28263.68) + 31.1
(28263.68) + ___
(28263.68)
35.5
25.6
)
-1
-2
2
55. (x + 2) - (x - 4)
x+2
x-2 -_
1
1
_
_
_
2 x-2
2
(x + 2)
x -4 x+2
(x
2)
(x
+
2)
__
-1
)
(x + 2) 2(x - 2)
-4
__
(x + 2) 2(x - 2)
(xy)
(
(
62.
2x 2 + 5x 3
_
64.
x-2
_
)
304
x
2
(
x 2x + 5x )
_
x
2
5x + 2x; x ≠ 0
x 2 - 2x - 48
63. __
x 2 + 10x + 24
(x - 8)(x + 6)
__
(x + 4)(x + 6)
x - 8 ; x ≠ -6 and x ≠ -4
_
x+4
x 2 - 3x + 2
x-2
__
(x - 1)(x - 2)
1 ; x ≠ 1 and x ≠ 2
_
x-1
Holt McDougal Algebra 2
3.
8-4 RATIONAL FUNCTIONS,
PAGES 592-599
:EROSATEACHREALVALUEOFxFORWHICH
px
CHECK IT OUT!
1a. g is f translated 4 units
left.
g
x
(OLESATANYPOINTWHERExb IFxb
ISAFACTOROFBOTHpANDqANDTHELINE
xbISNOTAVERTICALASYMPTOTE
IFDEGREEOFpDEGREEOFq
g
Ú
LEADINGCOEFFICIENTOFp
y ??
LEADINGCOEFFICIENTOFq
y
f
(ORIZONTALASYMPTOTESNONEIFDEGREE
OFpDEGREEOFqTHELINEyIF
DEGREEOFpDEGREEOFqTHELINE
b. g is f translated 1 unit
up.
y
6ERTICALASYMPTOTESATEACHREALVALUE
OFxFORWHICHqx
px
fx qx
f
EXERCISES
GUIDED PRACTICE
1. discontinuous
2. asymptotes: x = 3, y = -5;
D: {x | x ≠ 3};
R: {y | y ≠ -5}
3. zeros: -6, -1;
asymptote: x = -3
2. g is f translated 2 units
down.
y
4a. zeros: -5, 3;
asymptote: x = 1
x
4. g is f translated 1 unit right and 4 units up.
y
y
x
y
x
6. asymptotes: x = -4, y = 3;
D: {x | x ≠ -4};
R: {y | y ≠ 3}
5. hole at x = 2
7. asymptotes: x = 2, y = -8;
D: {x | x ≠ 2};
R: {y | y ≠ -8}
y
(OLE
ATx
x
5. asymptotes: x = 0, y = -1;
D: {x | x ≠ 0};
R: {y | y ≠ -1}
x
1 , 0;
c. zeros: -_
3
asymptotes: x = -3,
x = 3, y = 3
b. zero: 2;
asymptotes: x = -1,
x = 0, y = 0
x
x
y
x
y
3. g is f translated 5 units
left.
8. zeros: -3, 4
vertical asymptote:
x=0
9. zeros: 0, 5
vertical asymptote:
x=2
y
y
THINK AND DISCUSS
x
1. Possible answer: The x-value of the vertical
asymptotes are excluded from the domain.
x
2. Both types of functions may have more than one
zero. Rational functions may have one or more
asymptotes, but polynomial functions do not. The
domain of polynomial functions is all real numbers,
but the domain of many rational functions do not
include all real numbers.
305
Holt McDougal Algebra 2
20. asymptotes: x = -6, y = 0;
D: {x | x ≠ -6};
R: {y | y ≠ 0}
10. zero: 0;
11. zeros: -2, -1;
vertical asymptote: x = 1
asymptote: x = 3
y
y
x
x
12. zero: 2;
asymptotes: x = -6,
x = 0, y = 0
y
21. asymptotes: x = 0, y = 5;
D: {x | x ≠ 0};
R: {y | y ≠ 5}
22. asymptotes: x = 4, y = -1;
D: {x | x ≠ 4};
R: {y | y ≠ -1}
2;
13. zero: -_
5
asymptotes: x = -1,
y=5
23. zeros: -2, 5;
vertical asymptote:
x=2
y
14. hole at x = 3
y
x
25. zeros: -2, 2;
vertical asymptote:
x = -3
y
x
y
x
x
y
29. hole at x = 0
19. g is f vertically stretched by a factor of 2
x
y
(OLEAT
x
x
x
31. hole at x = 7
30. hole at x = 1
y
(OLEAT x
y
x
17. g is f translated 5 units 18. g is f translated 3 units
down.
left.
y
1 , 0;
28. zeros: -_
2
asymptotes: x = -1,
x = 1, y = -2
PRACTICE AND PROBLEM SOLVING
y
x
27. zero: 3;
asymptotes: x = -2,
x = 2, y = 0
y
(OLEAT
x
y
y
16. hole at x = -5
(OLEAT x
26. zeros: -1, 2;
vertical asymptote:
x=1
x
x
15. hole at x = 2
(OLEAT
x
x
x
x
x
y
y
24. zeros: -4, 2;
vertical asymptote:
x=1
y
(OLEAT
x
x
x
306
Holt McDougal Algebra 2
!VERAGECOST
2000 + 350
32a. f(x) = _
x
b.
42a.
x
b. D: ;
R: {y | 0 < y ≤ 3}
c. y = 0
2000 + 350 = 400
c. f(40) = _
y
(OLEAT
x
34. hole at x = 1
y
(OLEAT
x
x
36. zeros: -4, -2;
asymptotes: x = 0,
y=1
x
y
x
(x + 1)(x - 3)
39. Possible answer: f(x) = __
x
(x - 2)
_
40. Possible answer: f(x) =
x(x + 2)
x
)
40 + 15 = 40 + 225 = 265
c. xf(x) = 15 _
15
The total cost is $265.
45. Possible answer: The graph has only 1 vertical
asymptote, at x = -1. There is a hole at x = 1
because x - 1 is a factor of both the numerator and
denominator.
46. Possible answer: Yes; a rational function with a
denominator that is never equal to zero will have no
x - 3 ).
vertical asymptotes (for example: f(x) = _
x2 + 1
47a.
,APTIMES
38. zeros: ±3;
asymptotes: x = 2,
x = -2, y = 1
y
(
40 + k
b. f(x) = _
x
40 + k
55 = _
1
55 = 40 + k
k = 15
x
37. zero: 0;
asymptotes: x = 3,
x = -3, y = 0
$6$SPURCHASED
y
y
44a.
b. The chemist must add about 17 g salt.
3ALTADDEDG
2,
asymptotes: x = _
3
y = -2
x
5;
35. zero: _
6
!VERAGECOST
33. zero: -1;
asymptotes: x = 0,
y = 1;
hole at x = 3
#ONCENTRATION
43a.
40
The average cost per person is $400.
"ANDMEMBERS
y
(x - 2)(x + 3)
41. Possible answer: f(x) = __
(x + 3)(x + 1)
!VERAGESPEEDMIH
b. t = 12; the number of seconds the driver spent at
the pit stop
307
Holt McDougal Algebra 2
12(200) + 9000
c. t(200) = __ = 57
200
It takes 57 s in total.
48. The equation is false for x = -3 because the
denominator of the left side of the equation is equal
to 0 when x = -3. Division by 0 is undefined.
x+1
5x - 7 + _
3x - 6 64. _
x-1 -_
63. _
2x + 1
2x + 1
x+2
x-3
5x - 7 + 3x - 6
__
x+1 _
x+2
x-3 -_
x-1 _
_
2x + 1
x+2 x-3
x-3 x+2
8x - 13 ; x ≠ -_
1
_
(x - 1)(x - 3) - (x + 1)(x + 2)
___
2
2x + 1
(x + 2)(x - 3)
(
(
)
2
2
x - 4x + 3 - (x + 3x + 2)
___
49. Possible answer: The domain of a rational function
is all real numbers except the x-values of vertical
asymptotes and holes.
(x + 2)(x - 3)
-7x + 1
__
;
(x + 2)(x - 3)
x ≠ -2 and x ≠ 3
TEST PREP
50. A
51. F
8-5 SOLVING RATIONAL EQUATIONS
AND INEQUALITIES, PAGES 600-607
52. D
CHALLENGE AND EXTEND
2;
54. zeros: -3, -_
3
asymptotes: x = -2,
y
x = 2, y = 3;
(OLESATx
hole at x = 3
xANDx
53. holes at x = 1, x = 2,
x = 3;
y
(OLEAT
x
x
CHECK IT OUT!
10 = _
4
1a. _
3
x+2
10 (x) = _
4 (x) + 2(x)
_
x
3
10 x = 4 + 2x
_
3
_4 x = 4
3
x=3
6 -1
c. x = _
x
x
55a. Possible answer: 2
6 (x) - 1(x)
x(x) = _
x
2
x =6-x
x2 + x - 6 = 0
(x - 2)(x + 3) = 0
x - 2 = 0 or x + 3 = 0
x = 2 or x = -3
b. Possible answer: 1
c. Possible answer: 0
2
x +1
56. Possible answer: f(x) = _
x2 + 2
2
x(x - 1)
57. Possible answer: f(x) = __
(x 2 - 1)(x 2 - 9)
2a.
x 2 - 16
x-4
x-4
16 = 2(x + 4)
8=x+4
x=4
The solution x = 4 is extraneous.
Therefore there is no solution.
x 2 - 16
2,865,358 - 817,073
__
× 100% ≈ 250%
817,073
The percent increase in the number of females
participating is about 250%.
59. log 3 (5x - 2) = log 3 (2x + 8) 60. log 2 x 2 = 4
2
4
5x - 2 = 2x + 8
x =2
3x = 10
24
x = ± √
10
_
x=
x = ±4
3
1
_
=3
62. log 4 48 - log 4 4x = 4
61. log x
27
48 = 4
1
log 4 _
x3 = _
4x
27
3 _
12
_
1
log 4
=4
x=
x
27
12 = 4 4
_
1
x=_
x
3
12
x=_
44
3
x=_
64
√
16
2
_
=_
16 (x 2 - 16) = _
2 (x 2 - 16)
_
SPIRAL REVIEW
58.
5 = -_
6 +_
7
b. _
x
4
4
_6 = -3
x
_6 (x) = -3(x)
x
6 = -3x
x = -2
b.
1 =_
x +_
x
_
x-1
x-1
6
1 6(x - 1) = _
x 6(x - 1) + _
x 6(x - 1)
_
(x - 1)
(x - 1)
6 = 6x + x(x - 1)
6 = 6x + x 2 - x
2
x + 5x - 6 = 0
(x -1)(x + 6) = 0
x - 1 = 0 or x + 6 = 0
x = 1 or x = -6
The solution x = 1 is extraneous.
The only solution is x = -6.
( )
( )
308
(6)
Holt McDougal Algebra 2
)
2 +_
2
3. 5 = _
2-c
2+c
2
2 (4 - c 2) + _
2 (4 - c 2)
5(4 - c ) = _
2-c
2+c
THINK AND DISCUSS
1. Possible answer: The LCD is a multiple of each
denominator. Therefore, each denominator is a
factor of the LCD.
5(4 - c 2) = 2(2 + c) + 2(2 - c)
20 - 5c 2 = 8
-5c 2 = -12
c ≈ ±1.5
The average speed of the current is 1.5 mi/h.
2. Possible answer: When you multiply both sides
of an equation by a variable expression, you may
produce an equation with solutions that make
denominators of the original equation equal to 0.
1 (11) + _
1 (11) = 1
4. _
20
h
1 (11)(20h) + _
1 (11)(20h) = 1(20h)
_
20
h
11h + 220 = 20h
220 = 9h
24 ≈ h
It will take Remy about 24 min to mulch the garden
when working alone.
5a. 3 < x ≤ 4
6a.
4. Possible answer:
$EFINITION
EQUATIONSTHATCONTAINRATIONALEXPRESSIONS
%XAMPLES
x Ú
Ú
Ú
x
x
x x b. x = -5
y
3. Possible answer: (1) Graph each side of the
inequality. (2) Multiply both sides by x and consider
two cases, x is positive or x is negative.
y
x
#HARACTERISTICS
CANBESOLVEDBYMULTIPLYINGBOTHSIDES
BYTHE,#$OFALLTHETERMSINTHE
EQUATIONMAYGENERATEEXTRANEOUS
SOLUTIONSWHENSOLVED
2ATIONAL
%QUATIONS
.ONEXAMPLES
Ȗее
x ̎x̑ EXERCISES
GUIDED PRACTICE
x
1. Possible answer: An equation is a statement that
2 expressions are equal. A rational expression is
a quotient of 2 polynomials. A rational equation
contains at least 1 rational expression.
6 ≥ -4
_
x-2
LCD is positive.
6 (x - 2) ≥ -4(x - 2)
_
x-2
6 ≥ -4x + 8
4x ≥ 2
1;
x≥_
2
x-2>0
x>2
Solution in this case is
x > 2.
LCD is negative.
6 (x - 2) ≤ -4(x - 2)
_
x-2
6 ≤ -4x + 8
4x ≤ 2
1;
x≤_
2
x-2<0
x<2
Solution in this case is
1.
x≤_
2
1 or x > 2.
The solution to the inequality is x ≤ _
2
9 <6
b. _
x+3
LCD is negative.
LCD is positive.
9 (x + 3) > 6(x + 3)
9 (x + 3) < 6(x + 3)
_
_
x+3
x+3
9 < 6x + 18
9 > 6x + 18
-9 < 6x
-9 > 6x
3
3
x > -_;
x < -_;
2
2
x+3>0
x+3<0
x > -3
x < -3
The solution in this case
The solution in this case
3
is x < -3.
is x > -_.
2
3
The solution to the inequality is x < -3 or x > -_.
2
309
1 +_
2 =_
17
2. _
t
8t
8
17 (8t)
_1 (8t) + _2 (8t) = _
t
8t
8
t + 16 = 17
t=1
4.
1 -4
3. 7 = _
w
1 (w) - 4(w)
7(w) = _
w
7w = 1 - 4w
11w = 1
1
w=_
11
1 =_
7
_
r-5
2r
1 2r(r - 5) = _
7 2r(r - 5)
_
(r - 5)
5
1 =_
x -_
5. _
x
6
6
( 2r )
2r = 7(r - 5)
2r = 7r- 35
r=7
_1 (6x) = _x (6x) - _5 (6x)
x
6
6
2
6 = x - 5x
x 2 - 5x + 6 = 0
(x - 6)(x + 1) = 0
x - 6 = 0 or x + 1 = 0
x = 6 or x = - 1
1 =2
12 = 7
7. k + _
6. m + _
m
k
12 (m) = 7(m)
1 (k) = 2(k)
m(m) + _
k(k)
+_
m
k
2
m 2 + 12 = 7m
+ 1 = 2k
k
m 2 - 7m + 12 = 0
k 2 - 2k + 1 = 0
(m - 3)(m - 4) = 0
(k - 1) 2 = 0
m - 3 = 0 or m - 4 = 0
k-1=0
m = 3 or m = 4
k=1
Holt McDougal Algebra 2
-2x + _
4
x =_
8. _
3
x+2
x+2
-2x 3(x + 2) + _
4 3(x + 2)
x 3(x + 2) = _
_
3
x+2
x+2
-6x + x(x + 2) = 12
x 2 - 4x - 12 = 0
(x - 6)(x + 2) = 0
x = 6 or x = -2
The solution x = -2 is extraneous.
The only solution is x = 6.
(
9.
)
)
x(x + 1)
x2 + x
3
3
_
x(x + 1) - 1x(x + 1) = _
x(x + 1)
2
x(x + 1)
x +x
3 - x(x + 1) = 3
x(x + 1) = 0
x = 0 or x = -1
Both solutions are extraneous.
Therefore there is no solution.
(
16.
4 <4
_
17.
12 ≤ 3
_
18.
10 > 2
_
)
60 + _
60
11. 16.5 = _
8-c
8+c
2
60 (64 - c 2) + _
60 (64 - c 2)
16.5(64 - c ) = _
8-c
8+c
16.5(64 - c 2) = 60(8 + c) + 60(8 - c)
2
1056 - 16.5c = 960
-16.5c 2 = -96
c ≈ ±2.4
The average speed of the current is 2.4 mi/h.
This is close to 2 mi/h, so the barge would take
60 mi , or 10 h, to travel upstream
about _
8 - 2 mi/h
60 mi , or 6 h, to travel downstream.
and about _
8 + 2 mi/h
The trip should take about 16 h, which is close to
the given time, so the answer is reasonable.
1 (30) + _
1 (30) = 1
12. _
m
50
1 (30)(50m) + _
1 (30)(50m) = 1(50m)
_
m
50
30m + 1500 = 50m
1500 = 20m
75 = m
The job will take 75 min if the large copier is broken.
13. -5 < x < 0
14. x = -5
y
y
x
x
x
2
2x - 6
2x + x(x - 3) = 6x
2
x - 7x = 0
x(x - 7) = 0
x = 0 or x = 7
)
3
3
_
-1=_
(
y
6x
x +_
x =_
_
2
x-3
2x - 6
6x (2x - 6)
x (2x - 6) + _
x (2x - 6) = _
_
x-3
10.
(
()
15. x < 0 or x > 1
x+1
LCD is positive.
LCD is negative.
4 (x + 1) < 4(x + 1)
4 (x + 1) > 4(x + 1)
_
_
x+1
x+1
4 < 4x + 4
4 > 4x + 4
0 < 4x
0 > 4x
x > 0;
x < 0;
x+1>0
x+1<0
x > -1
x < -1
The solution in this case
The solution in this case
is x > 0.
is x < -1.
The solution to the inequality is x < -1 or x > 0.
x-4
LCD is negative.
LCD is positive.
12 (x - 4) ≥ 3(x - 4)
12 (x - 4) ≤ 3(x - 4)
_
_
x-4
x-4
12 ≤ 3x - 12
12 ≥ 3x - 12
24 ≤ 3x
24 ≥ 3x
x ≥ 8;
x ≤ 8;
x-4>0
x-4<0
x>4
x<4
The solution in this case
The solution in this case
is x ≥ 8.
is x < 4.
The solution to the inequality is x < 4 or x ≥ 8.
x+8
LCD is negative.
LCD is positive.
10 (x + 8) < 2(x + 8)
10 (x + 8) > 2(x + 8)
_
_
x+8
x+8
10 > 2x + 16
10 < 2x + 16
-6 > 2x
-6 < 2x
x < -3;
x > -3;
x+8>0
x+8<0
x > -8
x < -8
The solution in this case
No solution in this case.
is -8 < x < -3.
The solution to the inequality is -8 < x < -3.
PRACTICE AND PROBLEM SOLVING
10
1 =_
20.
19. 4 + _
x
2x
10 (2x)
1 (2x) = _
4(2x) + _
x
2x
8x + 2 = 10
8x = 8
x=1
310
n-3
_5 = _
n-4
4
5(n - 4) = 4(n - 3)
5n - 20 = 4n - 12
n=8
Holt McDougal Algebra 2
21.
1 =3
_
a-7
1 = 3(a - 7)
_1 = a - 7
3
22 = a
_
3
14 = 9 - z
23. _
z
3 =_
x
1 -_
22. _
x
4
4
_1 (4x) - _3 (4x) = _x (4x)
x
4
4
2
4 - 3x = x
2
x + 3x - 4 = 0
(x - 1)(x + 4) = 0
x = 1 or x = -4
1 (1.5) + _
1 (1.5) = 1
29. _
2
h
_1 (1.5)(2h) + _1 (1.5)(2h) = 1(2h)
2
h
1.5h + 3 = 2h
3 = 0.5h
6=h
It would take the apprentice 6 h.
30. 0 < x < 1
14 (z) = 9(z) - z(z)
_
z
14 = 9z - z
z - 9z + 14 = 0
(z - 2)(z - 7) = 0
z = 2 or z = 7
2
31. x = -3
y
y
2
4 =4
24. x + _
x
4 (x) = 4(x)
x(x) + _
x
2
x + 4 = 4x
x 2 - 4x + 4 = 0
(x - 2) 2 = 0
x -2 = 0
x=2
25.
(2)
(x - 3)
x+1
3x = 2x - 1
x = -1
The solution x = -1 is extraneous.
Therefore there is no solution.
x
y
x
1 <2
33. _
3x
LCD is positive.
1 (3x) < 2(3x)
_
3x
1 < 6x
1;
x>_
6
3x > 0
x>0
The solution in this case
1.
is x > _
6
LCD is negative.
1 (3x) > 2(3x)
_
3x
1 > 6x
1;
x<_
6
3x < 0
x<0
The solution in this case
is x < 0.
1.
The solution to the inequality is x < 0 or x > _
6
34.
2
2
_
=1+_
x-1
x(x - 1)
2
2 x(x- 1)
_
x(x - 1) = 1(x(x - 1)) + _
x-1
x(x - 1)
2 = x(x - 1) + 2x
x2 + x - 2 = 0
(x - 1)(x + 2) = 0
x = 1 or x = -2
The solution x = 1 is extraneous.
The only solution is x = -2.
)
3x = _
2x - 1
_
x+1
x+1
3x (x + 1) = _
2x - 1 (x + 1)
_
(
x
x+1
27.
8x + x(x - 3) = 24
x 2 + 5x - 24 = 0
(x - 3)(x + 8) = 0
x = 3 or x = -8
The solution x = 3 is extraneous.
The only solution is x = -8.
26.
32. 0 ≤ x < 5
4x + _
12
x =_
_
2
x-3
x-3
4x 2(x - 3) + _
x 2(x - 3) = _
12 2(x - 3)
_
(x - 3)
(
)
28. 22(550 - v) = 17(550 + v)
12100 - 22v = 9350 + 17v
2750 = 39v
71 ≈ v
The average speed of the wind is about 71 mi/h.
This is close to 70 mi/h. The plane travels about
22(550 - 70), or 10,560, mi on the flight to Bombay
and about 22(550 + 70), or 10,540, mi on the
flight to Los Angeles. Because the distances are
approximately equal, the answer is reasonable.
311
9 ≥ -6
_
x-4
LCD is negative.
LCD is positive.
9 (x - 4) ≤ -6(x - 4)
9 (x - 4) ≥ -6(x - 4) _
_
x-4
x-4
9 ≥ -6x + 24
9 ≤ -6x + 24
6x ≥ 15
6x ≤ 15
x ≥ 2.5;
x ≤ 2.5;
x-4>0
x-4<0
x>4
x<4
The solution in this case The solution in this case
is x > 4.
is x ≤ 2.5.
The solution to the inequality is x ≤ 2.5 or x > 4.
Holt McDougal Algebra 2
35.
9
_
>3
43.
x + 10
LCD is negative.
LCD is positive.
9
9 (x + 10) > 3(x + 10) _
_
(x + 10) < 3(x + 10)
x + 10
x + 10
9
> 3x + 30
9 < 3x + 30
-21
> 3x
-21 < 3x
x
< -7;
x > -7;
x + 10 > 0
x + 10 < 0
x > -10
x < -10
The solution in this case is No solution in this case.
-10 < x < -7.
The solution to the inequality is -10 < x < -7.
17,000
or equivalent inequality
(_
w )
36a. P ≤ 2w + 2
b. No; substituting 400 for P in the inequality results
in nonreal values of w.
37a. 2003
188 + h
191 ; 18 hits
b. Possible answer: _ = _
614
643
+
h
156 + h
0.5
c. _ = _
1
482 + h
156 + h = 241 + 0.5h
0.5h = 85
h = 170
170 hits; He had 326 more at bats than hits, and
this will stay constant with each hit. He would need
to reach a total of 326 hits from 156 hits, so the
answer is reasonable.
15n = _
5 -8
38. _
n-3
n-3
15n (n - 3) = _
5 (n - 3) - 8(n - 3)
_
n-3
n-3
15n = 5 - 8(n - 3)
15n = 5 - 8n + 24
23n = 29
29
n=_
23
z =_
z
39. _
z+1
z-4
z(z - 4) = z(z + 1)
z 2 - 4z = z 2 + z
-4z = z
z=0
4
1
_
40. + 6 = _
x
x2
1 (x 2)
_4 (x 2) + 6(x 2) = _
x
2
x
4x + 6x 2 = 1
6x 2 + 4x - 1 = 0
4 2 - 4(6)(-1)
-4 ± √
-2 ± √
10
x = __ = _
6
2(6)
8 -_
3 =_
6
41. _
x
x
x-1
6 x(x - 1)
_8 x(x - 1) - _3 x(x - 1) = _
x
x
x-1
8(x - 1) - 3(x - 1) = 6x
5x - 5 = 6x
x = -5
2(x + 4) _
3x
_
=
42.
x-4
x-4
2(x + 4) = 3x
2x + 8 = 3x
x=8
()
()
(
)
1 +_
4 =_
7
_
a-1
a+1
a2 - 1
1 (a 2 - 1) + _
4 (a 2 - 1) = _
7 ( a 2 - 1)
_
a-1
a+1
a2 - 1
(a + 1) + 4(a - 1) = 7
5a - 3 = 7
5a = 10
a=2
6 ≥_
5
44. _
r
2
LCD is negative.
LCD is positive.
_6 (r) ≥ _5 (r)
_6 (r) ≤ _5 (r)
r
r
2
2
6 ≥ 2.5r
6 ≤ 2.5r
r ≤ 2.4;
r ≥ 2.4;
The solution in this case
No solution in this case.
is 0 < r ≤ 2.4.
The solution to the inequality is 0 < r ≤ 2.4.
45.
8 >4
_
x+1
LCD is negative.
LCD is positive.
8 (x + 1) < 4(x + 1)
8 (x + 1) > 4(x + 1)
_
_
x+1
x+1
8 > 4x + 4
8 < 4x + 4
4 > 4x
4 < 4x
x < 1;
x > 1;
x+1>0
x+1<0
x > -1
x < -1
The solution in this case
No solution in this case.
is -1 < x < 1.
The solution to the inequality is -1 < x < 1.
4
46. x ≥ _
x
LCD is negative.
LCD is positive.
4 (x)
4
_
x(x) ≤ _
x(x) ≥ (x)
x
x
2
2
x ≥4
x ≤4
x ≤ -2 or x ≥ 2;
-2 ≤ x ≤ 2;
The solution in this case
The solution in this case
is x ≥ 2.
is -2 ≤ x < 0.
The solution to the inequality is -2 ≤ x < 0 or
x ≥ 2.
47. ±0.45
49. 0, 2
48. ±1.27
1 +_
7 =2
50. _
x
2
_1 = -_3
x
2
2
x = -_
3
500 ;
51a. 2001 winner: _
s
500
2002 winner: _
s + 25
312
Holt McDougal Algebra 2
500 + _
32 = _
500 ;
b. Possible answer: _
s
s + 25
60
15
(__
CHALLENGE AND EXTEND
) (
500s(s + 25) _
500s(s + 25)
8s(s + 25)
= 15 __
+
s
s + 25
15
)
58.
4x
7x
_
= __
59.
1 - 4x -1 + 3x -2 _
__
= x-1
7500s + 8s(s + 25) = 7500(s + 25)
8s 2 + 200s - 187500 = 0
200 2 - 4(8)(-187500)
-200 ± √
s = ___
2(8)
s ≈ 141 or -166
The average speed of the 2001 winner is 141 mi/h.
52. depending on the values of a, b, and c, either 2, 1,
or 0
(
)
()
x+3
-1
x-1
1-x
_
(_x ) = _
x
x+3
1 + 3x -1
x-1
x
1
_=_
x+3
x+3
all real numbers except -3, 0, and 3.
TEST PREP
()
1- 9x -2
(1 - 3x -1)(1 - x -1) _
__
= x-1
(1 - 3x -1)(1 + 3x -1) x + 3
(1 - x -1) _
_
= x-1
(1 + 3x -1) x + 3
53. Multiply each term by the LCD, 5x.
Divide out common factors, to get 3x 2 = 15 - 30x.
Simplify, to get x 2 + 10x - 5 = 0.
Use the quadratic formula to solve for x
x = -5 ± √
30 .
54. A;
3 =_
6
_1 + _
x
x
x+3
3 x(x + 3) = _
6 x(x + 3)
_1 x(x + 3) + _
x
x
x+3
(x + 3) + 3x = 6(x+ 3)
4x + 3 = 6x + 18
-15 = 2x
15
-_ = x
2
55. G;
x+2 _
4
_
- 1 =_
x
x-4
x 2 - 4x
x + 2( 2
1 (x 2 - 4x) = _
4
_
(x 2 - 4x)
x - 4x) - _
x
2
x-4
x - 4x
x(x + 2) - (x - 4) = 4
x2 + x = 0
x(x + 1) = 0
x = 0 or x + 1 = 0
x = 0 or x = -1
The solution x = 0 is extraneous.
The only solution is x = -1.
x2 + x - 6
x 2 - 5x - 24
4x(x
2)(x
+
3)(x - 8) __
7x(x - 2)(x + 3)(x - 8)
__
=
(x - 2)(x + 3)
(x - 8)(x + 3)
4x(x - 8) = 7x(x - 2)
2
3x + 18x = 0
3x(x + 6) = 0
x = 0 or x + 6 = 0
x = 0 or x = -6
60.
3x - _
2 ≥7
_
x+2
x+4
LCD is positive.
2(x + 2)(x + 4)
3x(x + 2)(x + 4) __
__
≥ 7(x + 2)(x + 4)
x+2
x+4
3x(x + 4) - 2(x + 2) ≥ 7(x + 2)(x + 4)
4x 2 + 32x + 60 ≤ 0
4(x + 3)(x + 5) ≤ 0
x + 3 ≤ 0 and x + 5 ≥ 0
-5 ≥ x ≤ -3;
(x + 2)(x + 4) > 0
x + 2 > 0 and x + 4 > 0
x > -2 or x < -4
The solution in this case is -5 ≤ x < -4.
LCD is negative.
3x(x + 2)(x + 4) __
2(x + 2)(x + 4)
__
≤ 7(x + 2)(x + 4)
x+2
x+4
3x(x + 4) - 2(x + 2) ≤ 7(x + 2)(x + 4)
4x 2 + 32x + 60 ≥ 0
4(x + 3)(x + 5) ≥ 0
x + 3 ≥ 0 and x + 5 ≥ 0
x ≥ -3 or x ≤ -5
(x + 2)(x + 4) < 0
x + 2 < 0 and x + 4 > 0
-4 < x < -2
The solution in this case is -3 ≤ x < -2.
The solution to the inequality is -5 ≤ x < -4 or
-3 ≤ x < -2.
56. B
1 (7) + _
1 (7) = 1
57a. Possible answer: _
15
h
1
1
_
_
b.
(7)(15h) + (7)(15h) = 1(15h)
15
h
7h + 105 = 15h
105 = 8h
13 ≈ h
It would take about 13 h to fill the tank.
313
Holt McDougal Algebra 2
61.
6 >_
x +5
_
4
x-3
LCD is positive.
6 4(x - 3) > _
x 4(x - 3) + (5)4(x - 3)
_
4
x-3
24 > x(x - 3) + 20(x - 3)
x 2 + 17x - 84 < 0
(x - 4)(x + 21) < 0
x - 4 < 0 and x + 21 > 0
-21 < x < 4;
x-3>0
x>3
The solution in this case is 3 < x < 4.
LCD is negative.
6 4(x - 3) < _
x 4(x - 3) + (5)4(x - 3)
_
x-3
4
24 < x(x - 3) + 20(x - 3)
2
x + 17x - 84 > 0
(x - 4)(x + 21) > 0
x - 4 > 0 and x + 21 > 0
x > 4 or x < -21;
x-3<0
x<3
The solution in this case is x < -21.
The solution to the inequality is x < -21 or
3 < x < 4.
(
)
(
)
-10
66. y = _
x
()
()
( )
65.
x
V1
V2
8.7 = _
m
_
1000
4500
1000m = 8.7(4500)
m = 39.15
The mass of the statue is 39.15 kg.
k
2. t = _
n
k
6=_
6
k = 36
36 = 3.6
k =_
t=_
n
10
It will take 10 workers 3.6 h to clean the rides.
3.
3
5x
_
x - 2x - 3
4. _
x 2 + 5x + 4
(x
- 3)(x + 1)
__
(x + 4)(x + 1)
x - 3;
_
x+4
x ≠ -4 and x ≠ -1
2
2
10x + 5x
3
5x
_
5x(2x + 1)
2
x
_
;
2x + 1
1 and x ≠ 0
x ≠ -_
2
5
_
4 √7
m
m1 _
_
= 2
1.
1 (4)(x)
63. 4(4 + x) - _
2
16 + 4x - 2x
2x + 16
√
8
√
√
-3
2
_3 _
√
8 √
2
-3 √
6
_
√16
6
-3 √
_
4
READY TO GO ON? PAGE 609
)
-3 √
3
_
x
5.
-x + 6
__
7.
9x 6y
x
_
÷_
SPIRAL REVIEW
64.
y
y
62. Let h be the number of hours needed for Marcus
to paint the barn individually, then 2h will be the
number of hours needed for Will.
1
1 (6) = _
_1 (6) + _
3
2h
h
1 (2h)
1 (6)(2h) = _
_1 (6)(2h) + _
3
2h
h
2h
12 + 6 = _
3
h = 27
2 ÷_
1 ÷_
1 =_
1 = 18
1-_
27
27
3
3
It will take Marcus 18 additional hours to complete.
(
3
67. y = _
x
( )
√
7
5 _
_
x - 3x - 18
-(x - 6)
__
(x - 6)(x + 3)
-1 ; x ≠ -3 and x ≠ 6
_
x+3
27x 2y 5
6y 2
2
4
6y
x ·_
_
x
4
3y
3
2x
_
y2
√
4 √7
7
5 √7
_
4(7)
5 √7
_
28
314
6.
2
x+3 _
_
· 2x - 4
x + 2 x2 - 9
2(x - 2)
x + 3 __
_
·
x + 2 (x + 3)(x - 3)
2(x - 2)
__
(x + 2)(x - 3)
2
x + x - 12
2x - 18x ÷ _
8. _
x 2 - 2x - 8
x 2 - 16
2
(x + 4)(x - 4)
2x(x - 9)
__
· __
(x - 4)(x + 2) (x + 4)(x - 3)
2x(x + 3)(x - 3) _
__
· x-4
x-3
(x - 4)(x + 2)
2x(x + 3)
_
x+2
3
Holt McDougal Algebra 2
9.
3x + 2 _
x+5
_
-
3
x -x +_
10. _
x+5
x 2 - 25
2
x-5
3 _
x -x +_
_
2
x+5 x-5
x - 25
2
x - x + 3(x - 5)
__
(x + 5)(x - 5)
2
x + 2x - 15
__
(x + 5)(x - 5)
(x + 5)(x - 3)
__
(x + 5)(x - 5)
x - 3 ; x ≠ -5 and x ≠ 5
_
x-5
2
x-2
x-2
3x + 2 - (x + 5)
__
x-2
2x - 3 ; x ≠ 2
_
x-2
(
10 = 3
17. y - _
y
10 (y) = 3(y)
y(y) - _
y
2
y - 10 = 3y
y 2 - 3y - 10 = 0
(y - 5)(y + 2) = 0
y - 5 = 0 or y + 2 = 0
y = 5 or y = -2
)
18.
x-8
x-8
x = 24 - 2x
x=8
The solution x = 8 is extraneous.
Therefore there is no solution.
x -_
1
11. _
x-3
x+3
(
x+3
x _
_
)
(
)
x-3
1 _
-_
x-3 x+3
x+3 x-3
x(x + 3) - (x - 3)
__
(x - 3)(x + 3)
2
x + 2x + 3
__
; x ≠ -3 and x ≠ 3
(x - 3)(x + 3)
12.
x + 15
-3x - _
19. _
=1
3
x+9
x + 15
-3x (x + 9) - _
_
(x + 9) = 1(x + 9)
3
x+9
-x(x + 9) - (x + 15) = x + 9
2
-x - 11x - 24 = 0
-(x + 3)(x + 8) = 0
x + 3 = 0 or x + 8 = 0
x = -3 or x = -8
2d
_
d +_
d
_
550
430
2d(23650)
___
d (23650) + _
d (23650)
_
550
47300d
_
43d + 55d
47300d ≈ 483
_
430
98d
The average speed for the entire trip is 483 mi/h.
13. g is f translated 4 units 14. g is f translated 1 unit
left and 2 units up.
right.
y
x
15. zeros: 4, -4;
vertical asymptote:
x=3
y
x
CHECK IT OUT!
1a. no real roots
16. zero: 0;
vertical asymptotes:
x = -2, x = 2,
horizontal asymptote:
y=0
8-6 RADICAL EXPRESSIONS AND
RATIONAL EXPONENTS,
PAGES 610-617
x
1 (3) + _
1 (3) = 1
20. _
4
h
_1 (3)(4h) + _1 (3)(4h) = 1(4h)
4
h
3h + 12 = 4h
h = 12
It would take the small oven 12 hours to bake
the bread.
y
x =_
24 - 2x
_
x-8
x-8
x (x - 8) = _
24 - 2x (x - 8)
_
c. 5
2a.
y
√4 16x 4
4
√
24 · x4
4
4
√
2 4 · √
x4
2 · x2
2x 2
b. ±1
√_x3
4
b.
8
√
x4 · x4
_
4
4
√
3
4
4
√x · √
x4
4
_
x
_
4
√
3
2
x
4
√
3
4
4
4
2
_
√
√
√
3
3 _
3
x ·_
_
·
·
4
4
4
4
√
√
3 √3
3 √
3
4
2√
3
x 3
_
√4 34
x 2 √
27
_
4
3
315
Holt McDougal Algebra 2
c.
_1
√x7 · √
x2
3
√
x7 · x2
3
√x9
3
3
√x
· x3 · x3
3
3
3
√x3 · √
x 3 · √
x3
3
3
3a. 64 3
2. Possible answer:
1
( √
64 )
3
(4)
4
0RODUCTOF0OWERS
1
?
x
u
?
x
?
x ?
u
?
_5
_3
b. 4 2
c. 625 4
(2)
32
4
5
?
xy
3
(5)
125
_3
b.
3
?
y
?
u
0OWEROFA1UOTIENT
?
?
?
?xy _3
√4 52
_1
1. 3
2. 3
3. ±5
4. 0
5.
5a. 36 8 · 36 8
_2
_3 + _1
54
36 8
_1
8
_1
52
√
8x 3
3
√
23 · x3
3
3
√
2 3 · √
x3
3
6.
_1
54
_9 - _1
(-8) 3
1
_
3
√
-8
1
_
-2
1
-_
2
54
52
25
) = 64(
?
?
?
32
√_
x
4
4
4
√
x4
4
√
24 · 2
_
4
√
x4
4
4
√
2 4 · √
2
_
4
4
√x
54
c. _
_1
?
√
25
_
_9
1
-_
4
2·x
2x
36 2
√
36
6
b. (-8) 3
1
_
?
x ?
?
?
y
GUIDED PRACTICE
10 3
10 3 = 1000
(
?
?
EXERCISES
√
10 9
_9
6. 64
Ú
x
3
n
-_
2 12
?
?
?
x
0OWEROFA0RODUCT
( √
625 )
5
4a. 81 4
c.
?
x x ?
?
0ROPERTIESOF2ATIONAL
%XPONENTS
x3
( √4)
1UOTIENTOF0OWERS
?
2
2 √
_
4
4
125x
√_
6
3
7.
6
x
8.
√
125x 6
_
3
3
√
6
3
3
√
5 · x3 · x3
)
2
x
2x · 5 · x
5x √2x
__
12
-_
2 12
√
50x 3
√
2x · 5 2 · x 2
2x · √
52 ·
3
√6
√
5 3 · √
x 3 · √
x3
__
3
3
= 64(2 )
1
= 64 _
2
= 32
The fret should be placed 32 cm from the bridge.
-1
3
3
√
6
()
2
5x
_
3
√
6
3
3
2
√
√
6 _
6
5x · _
_
· 3
3
3
√
√
√
6
6
6
23 2
6
5x √
THINK AND DISCUSS
_
1. Possible answer: When a and n are natural
n
a n can be written with a
numbers, the expression √
√
63
3
2
36
5x √
_
3
_n
rational exponent: a n . The exponent simplifies to 1,
so the expression becomes a 1, or a.
6
316
Holt McDougal Algebra 2
9.
4
√
x 8 · √
x4
3
4
√
x 4 · x 4 · √
x3 · x
3
4
4
3
√
x 4 · √
x 4 · √
x3 · √
x
3
( _)_
√_x4
5
3
10.
1 1
3
23. 64 2
_1 · _1
√
x3 · x2
_
64 2
√
4
3
√x 3 · √
x2
64 6
2
3
x · x · x · √x
3
x3 √
x
3
_
√
4
3
x √x2
3
√
4
3
√
x x2
√
2
_·_
3
3
√
√
4
2
3
2
x √2x
3
√
27
3
3
_1
26. 7 4 · 7
74
25 2
_1
5
1
_
_1
72
√
7
_
7
√
23
3
2x 2
x √
_
1
-_
2
√
-x 3
√
10 · 2 2 · x 2 · x 2
__
(-x) 3
√
√
10 · √
2 2 · √
x 2 · √
x2
__
3
-x
√
10 · 2 · x · x
__
-x
-2x √
10
12.
√_
x 12y 4
3
x4 · x4 · x4 · y4
√4 √
3
√
√
3 _
3 √
3
·_
· 4 ·_
4
4
√
√
√
√
3
3
3
3
4
x 3y √
33
4
4
√
34
x y √
27
_
4
3
_3
_3
4
PRACTICE AND PROBLEM SOLVING
30. -4
( √
36 )
3
5
_1
_2
16. 8 3
( √8 ) 2
3
(2) 2
4
3·x
3x
6
3 x
35. _
250
3
√
x3 · x3
√
_
3
√
250
3
3
√x · √
x3
3
_
_3
√
9 10
5
18. 8 2
3
√
250
2
x
_
3
√
250
3
2
√
4
x
_
_
·
3
3
√
250 √
4
23
√
4
x
_
3
√1000
23
4
x √
_
10
10
_
95
9 2 = 81
_3
_2
_1
19. 5 6 = 5 2
_1
20. 27 3
_3
21. 13 2 · 13 2
_1 + _3
13 2
13 2
169
2
_4
93
22. _
_2
93
_4 - _2
93
34.
√
33 · x3
3
3
√
3 3 · √
x3
3
(2) 3
8
15. (-27) 3
3
√
-27
-3
3
3
( √
32 )
3
31. 2
· √
3x 2
33. √9x
3
√
9x · 3x 2
14. 32 5
13. 36 2
17.
29. √
50 ≈ 3.68
3.68 ft = 3.68 × 12 in ≈ 44 in.
The side length of the cube is about 44 in.
32. no real roots
4
3
4
5
3
4
4
√
3
x 3y
_
_1 · 6
62
63
216
(-125) 3
4
x 3y
_
1 6
1
_
-5
1
-_
4
√
3
4
4
4
√x · √x 4 · √
x4 · √
y4
__
4
( _)
28. 6 2
_1
__
_
(6) 3
216
27. (-125) 3
1
_
4
3
4
1
-_
7 2
3
√
40x 4
_
3
-_
4
_1 - _3
_1
_
11.
3
_2
25. 25 2
1
_
3
_1
3
1
-_
_
8
(_
27 )
√8
_
3
_1
3
3
24.
3
√4 324x 8
√4 34 · 4 · x4 · x4
4
4
4
√4 · √
3 4 · √4
x 4 · √
x4
4
3 · √
4·x·x
4
3x 2 √
4
36.
5
x
_
45
√
x2 · x2 · x
__
√
45
√
x 2 · √
x 2 · √
x
__
√
45
2
√
x
x
_
√45
2
√
5
x √x _
_
·
√
√45
5
2
√
5x
x
_
√
225
2
√
5x
x
_
15
_2
93
3
√
92
3
3
√
81 or 3 √3
317
Holt McDougal Algebra 2
37.
4
3
√
x 10
√56
x9
38. _
4
3
√
x4
√
23 · 7 · x3 · x3 · x3
4 x 10
3
3
3
3
3
_
√
2 3 · √
7 · √
x 3 · √
x 3 · √
x3
4
_1
_3
( 81 )
49
_
_1
( 2 3) 4
4
2
4
8 4 or √
8
40.
√x7 · √
x6
5
√
x7 · x6
5
5
√x
· x5 · x3
5
5
5
√x5 · √
x 5 · √
x3
5
3
x · x · √x
2 5 3
x
x √
5
√
81
5
3
3
3
_1
56.
_4
( √
-1000 )
(-10) 4
10,000
√
14 3
3
6
4
4
47. ( √
144 )
2
3451 - 227 = 3224
A lion needs to consume about 3224 more Calories
each day than a house cat.
_2
_2
_2
2
( √8 ) 2 · ( √
64 )
3
2
(2) · (4)
4 · 16
64
√
144
1
_
12
_1
8
30
59
14.1 - 1.5 = 12.6
About 12.6 g more iodine-125 than iodine-131 will
remain after 30 days.
_1
27 3
1
1
_
30
_
( ) ≈ 1.5
_
1
≈ 14.1
Iodine-125: 20 × 100(_
2)
1
3 _
3
(2 3) 3
_
()
20
_
1
b. Iodine-131: 20 × 100 _
2
2
( )
144 2
_1
()
_t
1 h = 100 _
1 8 ≈ 18%
59a. 100 _
2
2
About 18% of the sample will remain.
3
2
51. _
27
2
b. House cat: P = 73.3 √
4.5 3 ≈ 227
4
170 3 ≈ 3451
Lion: P = 73.3 √
8 3 · 64 3
1
-_
_1
( _)
_1
50. 144 2
1
_
2
4
58a. P = 73.3 √
55 3 ≈ 1480
A cheetah has a metabolism rate of 1480 Cal/day.
14 3
14 1
14
144 2 = _212
49. (8 · 64) 3
_3
_1
_1
57. a(6) = 1000 2 24 ≈ 1189
The amount will be about $1189 after 6 years.
144 4
48. 48 2
9
92
3
4
_3
46. (-8) 5
5 9 or √
5
_1
3
45.
6 √6
( )
27
(_
3)
43. (-1000) 3
_3
3
6
27
_
27 3
2
√
6 2 · √6
4
12 2
_4
44. 6 2
√
63
12 4
_1
√3
1 or _
_
_1
41. 64 2
√
64
8
3
_2
( √
216 )
(6) 2
36
53
1
-_
12 2
·x·x·x·y
(-3) · √2
3 3
√
-3x y 2
3
_1 · _1
12 4
_1 - _3
3
42. 216 3
1 1
3
55. 5 3
_3
3
3
( _)_
12 4
54. _
3
3
7
_1
x 9y 3
√-54
(-3) 3 · 2 · x 3 · x 3 · x 3 · y 3
√
· √
(-3) 3 · √2
x 3 · √
x 3 · √
x 3 · √
y3
√
2
1
_
_7
9
_9
x √
x
39.
_1
1
_
√
49
_
_1
√x · x
4
4
√
x 4 · √
x2
4
( )
4
24
x
4
√
x6
3
_1
49 - 2
53. _
81
1
_
_1 + _1
22
√
·x·x·x
2 · √7
3
2x 3 √
7
_1
52. 2 2 · 2 4
4
_
60. W = 35.74 + 0.6215(40) - 35.75(35) 25 +
4
_
2
_
0.4275(40)(35) 25 ≈ 28
The wind chill is about 28°F.
√
27
3
_2
3
318
Holt McDougal Algebra 2
_1
2π √Lg
g
_
√
L _
2π √
2π √L
L =_
=
·
=_
61a. 2π _
g
g
g
√g
√g
√
√
74. Never; possible answer: if x is positive, (x) 3 is
_1
positive and (-x) 3 is negative; therefore,
(0.35)(9.8)
2π √
b. t = __ ≈ 1.2
9.8
It will take about 1.2 s to complete one swing.
62.
√
20x 3
)
1
_
63. (5x) 2
4
_1
20 4 (x 3) 4
_1 _3
3
-9 √
x)
64. ( √
5
4
65.
5
6
( √
11x 8 )
4
_6
4
4
( √
-9 ) ( √x)
(11x 8) 4
3
_4 _4
_6
(-9) 5 x 3
11 4
76. Sometimes; possible answer: if x is positive,
3
3
-√
x is less than 0. If x is negative, - √
x is greater
than 0. Therefore, the inequality is true only for
positive values of x.
_6
(x 8) 4
_3
11 2 x 12
_2
12
66. (-8x ) 3
_7 _4
_2
_2
77. 2 and 3; about 2.62
67. 5 4 x 3
_3
51 · 5 4 · x 4
(
5 · (5x) 4
4
)3
5 √(5x
)
2 8
(-2) x
4x 8
_3
_3
_1
_3
_1
(-12) 3
√
5
_1
_1
_1
( a 2 b) 3 · b
3
a 2b
b √
5
9
54
-2x _1√
4
a
_
4
70.
b
82. 4 values; 1, 2, 3, 6
1 power, or use
83. Possible answer: Raise 10 to the _
6
x
the √ function to find the sixth root of 10.
5
_3
71. a 4 (4b
_1
6 4
_1
)
_1
(a 3) 4 (4b 2 · b 4) 4
_1
(a 4) 4
_
_1
_1
_1
(a 3) 4 · (4b 2) 4 · (b 4) 4
_1
b4
_1
(4a 3b 2) 4 · b 1
4
4a 3b 2
b √
1
a
_
4
√
b
4
4
4
_
√
√
√
b _
b
b
a ·_
_
·
·
4
4
4
4
√b
√
b √
b √b
4
3
√
a b
84. Possible answer: The exponent 2.4 is rational
because it can be expressed as the ratio of
2 integers. For example, the exponent can be
expressed as the ratio of the integers 24 and 10
24 .
because 2.4 = _
10
TEST PREP
85. A
86. H;
_1
_2
(4π) 3 (3(8V)) 3
_
_1
_2
b
_2
8 3 (4π) 3 (3V) 3
24
_
_1
72. m(24) = 100 · 2 60 ≈ 132
The mass is about 132 kg after 24 h.
73. Always; possible answer:
numbers.
_4
incorrectly applied. To simplify 625 3 ÷ 625 3 , you
should subtract the exponents rather than dividing
them.
(a 2b) 3 · (b 3) 3
5
( )
81. A is incorrect. The Quotient of Powers Property is
_1
(a 2b · b 3) 3
(x 15) 5
5
3
69. (a 2b 4) 3
(-2) 6 · (-3) 3
√
x 9 √
(-2) 5 · √
(-2)(-3) 3
9
b. P = 14.7(10) -0.000014(29028) ≈ 5.8
The air pressure at the top of Mount Everest is
about 5.8 psi.
125x
5 √
4
15
68. (-12x ) 5
x
80a. P = 14.7(10) -0.000014(5280) ≈ 12.4
The air pressure in Denver is about 12.4 psi.
_3
_2
(-2) 3 3 x 8
78. 3 and 4; about 3.76
79. -5 and -4; about -4.31
_3
(-8) 3 (x 12) 3
9
_8
x2
1
_
is positive for all nonzero real
The expression
x2
4
8
-2
numbers. Therefore, - √x ≠ x .
20 4 x 4
x
_1
x 8 = -x 4 = -x 2. The expression
equation: - √
-x 2 is negative for all nonzero real numbers.
-2
1.
Simplify the right side of the equation: x = _
20x 3 4
(-12) 5
_1
75. Never; possible answer: simplify the left side of the
_1
_1
_1
1
_
and (-x) 3 is positive; therefore, (x) 3 ≠ (-x) 3 .
_7
4
(
_1
(x) 3 ≠ (-x) 3 . If x is negative, (x) 3 is negative
_2
4(4π) 3 (3V) 3 = 4S
_6
√
x 6 = x 3 = x 2 for all real
3
319
Holt McDougal Algebra 2
87. A;
4
√
a
=
=
3;
100. zero: _
4
asymptotes: x = -6, y = 4
88. F;
6
3 56a
_
√
√
x6
4
_6
7
√8a 6
3
√
23 · a3 · a3
x4
3
_3
√
x3
= x2 =
y
√
2 3 · √
a 3 · √
a3
3
3
3
2a 2
x
CHALLENGE AND EXTEND
89.
((20_)_)
_1
1 1
2
2 2
= 20
(_12 )
_1
3
_1
_1
90. 2 3 · 4 6 · 8 9
_1
_1
_1
8-7 RADICAL FUNCTIONS,
PAGES 619-627
_1
2 3 · (2 2) 6 · (2 3) 9
= 20 8
_1
_1
CHECK IT OUT!
_1
23 · 23 · 23
1a. D: {x | x ∈ };
R: {y | y ∈ }
_1 · 3
23
21 = 2
b. D: {x | x ≥ -1};
R: {y | y ≥ 0}
y
y
3
a
91. √
>a
a is positive.
3
a2
1 > √
x
a is negative.
3
1 < √
a2
2a. g is f translated 1 unit
up.
g
y
92. 1 = 1 · 1 · 1 = 1;
3
(-1 + i √3)(-1 + i √3)(-1 + i √3 )
-1 + i √3
_
= ___
8
2
)(-1 + i √3
) _
(___
-2 - 2i √3
= 8 = 1;
=
8
8
3
(-1 - i √3)(-1 - i √3)(-1 - i √3 )
-1 - i √3
___
_
=
8
2
(
)
(-2 + 2i √3 )(-1 - i √3) 8
= ___ = _
=1
8
8
⎢
⎣ 9 5
95. ⎡ 6
⎢
⎢
5 ⎦
⎣ 0
y
g
f
x
y
f
x
⎣ 0 -6 ⎦
g
4. g(x) = -2 √
x+1
1 (x - 1) 2
96. g(x) = _
25 = _
256 x ;
64x ÷ _
√
√
4
25
256
_
h(50) = √
(50) ≈ 23
25
2
97. h(x) = -x + 3
3
98. zeros: ±2;
asymptote: x = -5
5. h(x) =
99. zero: -3;
asymptotes: x = -5,
x = -1, y = 0
y
The downward velocity is about 23 ft/s.
y
x
2 ⎦
x
b. g is f vertically stretched
by a factor of 3,
reflected across the
x-axis, and translated
1 unit down.
6 ⎤
f
g
3a. g is f reflected across
the y-axis and
translated 3 units up.
94. ⎡ -4 -8 ⎤
y
SPIRAL REVIEW
93. ⎡ 7 7 -2 ⎤
x
b. g is f vertically
compressed by a factor
1.
of _
2
f
3
)
x
1 > a2
1 < a2
-1 < a < 1
a < -1 or a > 1
The solution in this case
The solution in this case
is 0 < a < 1.
is a < -1.
The solution to the inequality is a < -1 or
0 < a < 1.
(
6a.
b.
y
y
x
x
x
320
Holt McDougal Algebra 2
4. D: {x | x ≥ 3};
R: {y | y ≥ 0}
THINK AND DISCUSS
1. Possible answer: No; an asymptote is a line that
a curve approaches as ⎪x⎥ or ⎪y⎥ becomes very
large. There is no such line in the graphs of radical
functions.
g x 6. D: {x | x ∈ };
R: {y | y ∈ }
y
x
x
x g x Ȗее
y
y
f
x
8. g is f translated 7 units
down.
(ORIZONTALTRANSLATION
y
x
7. D: {x | x ∈ };
R: {y | y ∈ }
y
'RAPH
Ȗе
x x
6ERTICALTRANSLATION
y
x
3. Possible answer:
%QUATION
y
2. Possible answer: The radicand must be
nonnegative, so 2x + 2 must be greater than or
equal to 0. Solve this inequality for x.
2x + 2 ≥ 0
x ≥ -1
The domain of the function is x ≥ -1.
4RANSFORMATION
5. D: {x | x ∈ };
R: {y | y ∈ }
x
9. h is f vertically stretched
by a factor of 3.
y
h
g
f
g x Ȗxе
2EFLECTION
y
x
x
10. j is f translated 5 units
right.
6ERTICALSTRETCH
g x Ȗеx
y
y
f
j x
x
f
y
11. g is f compressed
vertically by a factor
1 and translated
of _
2
1 unit down.
g
x
EXERCISES
GUIDED PRACTICE
1. Possible answer: The function rule is a radical
expression that contains a variable in the radicand.
2. D: {x | x ≥ -6};
R: {y | y ≥ 0}
3. D: {x | x ≥ 0};
R: {y | y ≥ -1}
12. h is f stretched
13. j is f reflected across
horizontally by a factor
the y-axis and then
of 3 and translated
translated 3 units right.
y
4 units left.
x
x
j
x
x
f
f
h
y
y
y
321
Holt McDougal Algebra 2
14. g is f reflected across 15. h is f reflected across
the x-axis, vertically
the y-axis, horizontally
stretched by a factor of
compressed by a factor
1 , and then
2, and then translated
of _
2
4 units down.
translated 2 units left.
y
f
x
h
y
x
x
28. D: {x | x ∈ };
R: {y | y ∈ }
29. D: {x | x ∈ };
R: {y | y ∈ }
y
g
y
f
27. D: {x | x ∈ };
R: {y | y ∈ }
y
x
26. D: {x | x ≥ -1};
R: {y | y ≥ -3}
x
y
16. j is f vertically stretched by a factor of 3 and then
translated 3 units up and 3 units left.
x
y
j
30. g is f translated 2 units 31. h is f translated 4 units
up.
right.
f
y
x
g
f
17. g(x) = 4 √
(x + 5) - 2
18. g(x) =
√
√
y
j
x
x
25. D: {x | x ≥ 0};
R: {y | y ≤ 0}
f
x
f
x
x
x
y
y
j
h
PRACTICE AND PROBLEM SOLVING
f
y
24. D: {x | x ≥ 2};
R: {y | y ≥ 0}
g
34. h is f vertically
35. j is f translated 4 units
compressed by a factor
left and 1 unit down.
y
1
_
of and then reflected
4
across the y-axis.
y
y
x
x
x
23.
y
22.
x
f
x
32. j is f vertically
33. g is f horizontally
compressed by a factor
compressed by a factor
1 and then translated
of 0.5.
of _
y
3
5 units left.
21.
y
h
x
6 _5 x ;
19. g(x) = _
5 9
6 _
5 (6) ≈ 2.2
g(6) = _
5 9
The distance to the horizon is about 2.2 mi.
f
- 7)
√-2(x
20.
y
322
Holt McDougal Algebra 2
36. g is f reflected across 37. h is f reflected across
the x-axis, vertically
the y-axis, vertically
stretched by a factor of
stretched by a factor of
4, and then translated
3, and then translated
1 unit up.
2 units up.
y
f
h
x
11000 ≈ 373
55a. D(11000) = 3.56 √
The approximate distance is 373 km.
b. A = 11000 - 4000 = 7000
D(7000) = 3.56 √
7000 ≈ 297
373 - 297 = 76
It will appear to decrease by about 76 km.
y
56. The speed is about 150 mi/h.
f
g
3PEEDMIH
x
38. j is f reflected across the y-axis, vertically
1 , and then translated
compressed by a factor of _
3
2 units left.
y
$EPTHFT
j
57. Yes; M j = 100
f
M > M j.
x
4909(32)(3960) ≈ 24941.3
√
v Moon = √4909(1/6)(32)(1080) ≈ 5317.5
40. g(x) =
1 (x - 2)
-_
√
6
24941.3 - 5317.5 ≈ 19624
The vehicle need to travel about 19,624 mi/h faster
on Earth than on the Moon.
41. g(x) = - √
x+1-4
3 = _
x ÷_
x ;
_
√
√
5
40
24
216 = 3
g(216) = √_
24
59a.
42. g(x) =
0ERIODS
The radius of the can is 3 cm.
43.
44.
y
y
45.
x
x
1
-_
b. Possible answer: The graph of T is a vertical
stretch of the graph of f by a factor of 2π and
a horizontal stretch of the graph of f by a factor
of 9.8.
y
46.
y
,ENGTHM
x
x
(-263 + 273) 3
= 100 and so
1000
58. v Earth =
1 √
39. g(x) = _
x+3
3
√__
47a. h(0.01) = 241(0.01) 4 ≈ 762
The resting heart rate is about 762 beats/min.
c. by a factor of 4
60. sometimes
61. sometimes
62. always
63. never
64. no
65. yes
y
y
1
-_
241(300) 4
b. h(300) =
≈ 58
The resting heart rate is about 58 beats/min.
48. a vertical stretch by a factor of 6 followed by a
translation 1 unit left
x
53. A
54. C
x
66. yes
y
x
50. a reflection across the x-axis followed by a
translation 3 units right and 7 units down
52. B
49. a vertical stretch by a factor of 3 followed by a
translation 1 unit right and 9 units down
51. D
323
Holt McDougal Algebra 2
TEST PREP
3PEEDOFSOUNDINAIRMS
67a.
73. D
75. A
76. J
77. Possible answer: The graph was reflected across
the y-axis and then translated 4 units right;
g(x) = √
-(x - 4) .
74. J
CHALLENGE AND EXTEND
√
1 (x + 3) + 4
78. g(x) = - √
x - 3 + 2 79. f(x) = -2 _
5
or equivalent function
4EMPERATUREƒ#
b. The speed of sound at 25˚C is about 346 m/s.
c. Possible answer: Because k is positive, the value
of the function is 0 only when T + 273.15 = 0. The
value of T that makes this equation true is
-273.15˚C.
SPIRAL REVIEW
80. -4x + 5 < -7
-4x < -12
x>3
68a. s is f translated 1 unit up.
y
x
b. s(45) = √
45 + 1 ≈ 7.7
8 samples should be taken.
240 ≈ 3.9
√_
4
100
_
= 2.5
=√
4
69. t 1 =
t2
3.9 - 2.5 = 1.4
it will take about 1.4 s longer.
70. Possible answer: A vertical compression of the
1 can be represented
parent function by a factor of _
2
1
_
√
by g(x) =
x . A horizontal stretch of the parent
2
function by a factor of 4 can be represented by
1 x . The expression _1 x can be simplified
h(x) = _
4
4
1
1
1 · √x . Therefore,
_
_
as follows:
x=
· √
x=_
4
4
2
1 √x .
g(x) = h(x) = _
2
71. Possible answer: Only nonnegative radicands have
real square roots; therefore, the domain of squareroot functions is limited to values of x that make the
radicand nonnegative. By contrast, all real radicands
have cube roots; therefore, the domain of cube-root
functions is not limited.
√
√
√
82. 2(x + 1) ≥ x - 2
2x + 2 ≥ x - 2
x ≥ -4
83. y = 2x - 10 1
2x + y = 14 2
Substitute equation 1 into equation 2.
2x + (2x - 10) = 14
4x = 24
x=6
Substitute x = 6 into equation 2.
y = 2(6) - 10
y=2
81. 12 ≥ 4(x - 5)
3≥x-5
x≤8
√
72. Possible answer: A horizontal translation affects
the domain, but not the range. The domain of the
translated function is x ≥ h, where h is the number
of units the function is translated horizontally.
A vertical translation affects the range, but not the
domain. The range of the translated function is
f(x) ≥ k, where k is the number of units the function
is translated vertically.
84. y = 3x - 2 1
3x = 2y 2
Substitute equation 1 into equation 2.
3x = 2(3x - 2)
3x = 6x - 4
4
x=_
3
4 into equation 1.
Substitute x = _
3
4 -2
y=3 _
3
y=2
()
85. -8x + y = 36 1
y=x-4 2
Substitute equation 2 into equation 1.
-8x + (x - 4) = 36
-7x = 40
40
x = -_
7
40 into equation 2.
Substitute x = -_
7
40
y = -_ - 4
7
68
y = -_
7
324
Holt McDougal Algebra 2
16
7 +x=_
86. _
x
3
16 (3x)
_7 (3x) + x(3x) = _
x
3
2
21 + 3x = 16x
3x 2 - 16x + 21 = 0
(x - 3)(3x - 7) = 0
x - 3 = 0 or 3x - 7 = 0
7
x = 3 or x = _
3
9
2 =_
87. 4 + _
x
2
_2 = _1
x
2
2
x=_
_1
2
x=4
_1
4a. (x + 5) 3 = 3
3
√
x+5=3
3
( √
x + 5) = 33
_1
b. (2x + 15) 2 = x
√
2x + 15 = x
2
( √
2x + 15 ) = (x) 2
3
x + 5 = 27
x = 22
2
2x + 15 = x
0 = x 2 - 2x - 15
0 = (x - 5)(x + 3)
x - 5 = 0 or x + 3 = 0
x = 5 or x = -3 ✓
_1
2
-5x = _
3x - 2
88. _
5a. √
x-3+2≤5
√
x-3≤3
c. 3(x + 6) 2 = 9
3 √
x+6
=9
x+5
x+5
2
-5x = 3x - 2
2
5x + 3x - 2 = 0
(x + 1)(5x - 2) = 0
x + 1 = 0 or 5x - 2 = 0
2
x = -1 or x = _
5
2
(3 √
x + 6) = 92
2
( √
x - 3 ) ≤ (3) 2
x-3≤9
x ≤ 12;
x-3≥0
x ≥ 3;
3 ≤ x ≤ 12
9(x + 6) = 81
x+6=9
x=3
3
b. √
x+2≥1
3
( √
x + 2) ≥ 13
8-8 SOLVING RADICAL EQUATIONS AND
INEQUALITIES, PAGES 628-635
3
x+2≥1
x ≥ -1
CHECK IT OUT!
b. √
3x - 4 = 2
1a. 4 + √
x-1=5
√
x-1=1
( √
x - 1)
2
=1
x-1=1
x=2
3
( √
3x - 4 )
3
2
c. 6 √
x + 10 = 42
√
x + 10 = 7
3
6. Possible answer:
s = √30fd
30 = √
30(0.7)d
30 = √21d
2
21d 2
(30) = √
3
=2
3x - 4 = 8
3x = 12
x=4
2a. √
8x + 6 = 3 √x
2
( √
x + 10 ) = 7 2
900 = 21d
43 ≈ d
If the car were traveling 30 mi/h, its skid marks
would have measured about 43 ft. Because the
actual skid marks measure less than 43 ft, the car
was not speeding.
2
( √
8x + 6 ) = (3 √
x)2
x + 10 = 49
x = 39
8x + 6 = 9x
x=6
3
b. √
x + 6 = 2 √
x-1
3
3
3
( √
x + 6 ) = (2 √
x - 1)
3
3
THINK AND DISCUSS
x + 6 = 8(x - 1)
14 = 7x
x=2
1. Possible answer: The equation can be solved
algebraically by squaring both sides, or it can be
solved by graphing both sides of the equation.
3a. √
2x + 14 = x + 3
2. Possible answer: To solve x 2 = a, take the square
root of each side.
To solve √
x = b, square each side. The operations
used to solve each equation are inverses of each
other.
2
( √
2x + 14 ) = (x + 3) 2
2
2x + 14 = x + 6x + 9
0 = x 2 + 4x - 5
0 = (x - 1)(x + 5)
x - 1 = 0 or x + 5 = 0
x = 1 or x = -5
x = -5 is extraneous, the only solution is x = 1.
3. Possible answer:
b. √
-9x + 28 = -x + 4
)SOLATE
THERADICAL
EXPRESSION
2
( √
-9x + 28 ) = (-x + 4) 2
2
-9x + 28 = x - 8x + 16
0 = x 2 + x - 12
0 = (x - 3)(x + 4)
x - 3 = 0 or x + 4 = 0
x = 3 or x = -4
325
2AISEBOTH
SIDESOFTHE
EQUATIONTO
THEPOWER
EQUALTOTHE
INDEXOFTHE
RADICAL
3IMPLIFY
ANDSOLVE
#HECK
SOLUTIONSIN
ORIGINAL
EQUATION
A)FTRUETHE
SOLUTIONISA
SOLUTIONOFTHE
RADICALEQUATION
B)FFALSETHE
SOLUTIONIS
EXTRANEOUS
Holt McDougal Algebra 2
x + 6 = x 2 + 8x + 16
0 = x 2 + 7x + 10
0 = (x + 2)(x + 5)
x + 2 = 0 or x + 5 = 0
x = -2 (x + 4 ≥ 0) ✓
EXERCISES
GUIDED PRACTICE
1. No; the expression under the radical does not
contain a variable.
2. √
x-9=5
( √
x - 9)
3. √
3x = 6
2
=5
x - 9 = 25
x = 34
-x - 1 = x 2 + 2x + 1
0 = x 2 + 3x + 2
0 = (x + 1)(x + 2)
x + 1 = 0 or x + 2 = 0
x = -1 (x + 1 ≥ 0) ✓
5. √
3x - 1 = √
2x + 4
3
3
( √
x - 2) = 23
2
2
( √
3x - 1 ) = ( √
2x + 4 )
3
x-2=8
x = 10
3x - 1 = 2x + 4
x=5
6. 2 √
x = √
x+9
5
7. √
x + 4 = √
3x - 2
16. √
15x + 10 = 2x + 3
5
(2 √x) 2 = ( √
x + 9)
5
5
( √
x + 4 ) = ( √
3x - 2 )
2
5
4x = x + 9
3x = 9
x=3
(2 √x) = ( √
x + 7)
8x = x + 7
7x = 7
x=1
3
5
x + 4 = 3x - 2
6 = 2x
x=3
3
3
8. 2 √
x = √
x+7
3
2
( √
-x - 1 ) = (x + 1) 2
3x = 36
x = 12
4. √x
-2=2
3
15. √
-x - 1 = x + 1
2
( √
3x ) = 6 2
2
3
9. √
x + 6 - √
2x - 4 = 0
√
x + 6 = √
2x - 4
2
2
( √
x + 6 ) = ( √
2x - 4 )
x + 6 = 2x - 4
x = 10
2
( √
15x + 10 ) = (2x + 3) 2
2
15x + 10 = 4x + 12x + 9
0 = 4x 2 - 3x - 1
0 = (x - 1)(4x + 1)
x - 1 = 0 or 4x + 1 = 0
1
x = 1 or x = -_
4
_1
_1
18. (2x + 1) 3 = 2
17. (x - 5) 2 = 3
3
√
√
x-5=3
2x + 1 = 2
2
( √
x - 5) = 32
(4 √
x + 1)
= (3 √
x + 2)
16(x + 1) = 9(x + 2)
7x = 2
2
x=_
7
2
2
_1
2
( √
4x + 5 ) = x 2
2
4x + 5 = x
0 = x 2 - 4x - 5
0 = (x - 5)(x + 1)
x - 5 = 0 or x + 1 = 0
x = 5 (x ≥ 0) ✓
2
( √
x + 56 ) = x 2
2
12. √
x + 18 = x - 2
_1
20. 2(x - 50) 3 = -10
3
2 √
x - 50 = -10
3
(2 √
x - 50 ) = (-10) 3
3
2
( √
x + 18 ) = (x - 2) 2
2
x + 18 = x - 4x + 4
0 = x 2 - 5x - 14
0 = (x - 7)(x + 2)
x - 7 = 0 or x + 2 = 0
x = 7 (x - 2 ≥ 0) ✓
13. √
3x - 11 = x - 3
2x + 1 = 8
2x = 7
7
x=_
2
19. (4x + 5) 2 = x
√
4x + 5 = x
11. √
x + 56 = x
x + 56 = x
0 = x 2 - x - 56
0 = (x - 8)(x + 7)
x - 8 = 0 or x + 7 = 0
x=8✓
3
x-5=9
x = 14
10. 4 √
x + 1 = 3 √
x+2
3
( √
2x + 1 ) = 2 3
8(x - 50) = -1000
x - 50 = -125
x = -75
_1
21. 2(x + 1) 2 = 1
2 √
x+1=1
2
(2 √
x + 1) = 12
4(x + 1) = 1
4x = -3
3
x = -_
4
_1
23. √
x+5-1≤4
√
x+5≤5
22. (45 - 9x) 2 = x - 5
√
45 - 9x = x - 5
2
( √
45 - 9x ) = (x - 5) 2
2
( √
3x - 11 ) = (x - 3) 2
45 - 9x = x 2 - 10x + 25
0 = x 2 - x - 20
0 = (x - 5)(x + 4)
x - 5 = 0 or x + 4 = 0
x = 5 (x - 5 ≥ 0) ✓
3x - 11 = x 2 - 6x + 9
0 = x 2 - 9x + 20
0 = (x - 4)(x - 5)
x - 4 = 0 or x - 5 = 0
x = 4 or x = 5
2
( √
x + 5) ≤ 52
x + 5 ≤ 25
x ≤ 20;
x+5≥0
x ≥ -5;
-5 ≤ x ≤ 20
14. √
x+6-x=4
√
x+6=x+4
2
( √
x + 6 ) = (x + 4) 2
326
Holt McDougal Algebra 2
25. √
2x + 5 < 5
24. √
2x + 6 ≤ 10
√
2x ≤ 4
2
( √
2x ) ≤ 4 2
2
( √
3x + 13 ) = (2x - 3) 2
2x + 5 < 25
2x < 20
x < 10;
2x + 5 ≥ 0
2x ≥ -5
5
x ≥ -_;
2
5
-_ ≤ x < 10
2
2x ≤ 16
x ≤ 8;
2x ≥ 0
x ≥ 0;
0≤x≤8
26.
37. √
3x + 13 + 3 = 2x
√
3x + 13 = 2x - 3
2
( √
2x + 5 ) < 5 2
2
3x + 13 = 4x - 12x + 9
0 = 4x 2 - 15x - 4
0 = (x - 4)(4x + 1)
x - 4 = 0 or 4x + 1 = 0
x = 4 (2x ≥ 0) ✓
_1
39. (x - 9) 2 = 4
38. √
x + 8 - x = -4
√
x+8=x-4
√
x-9=4
2
( √
x + 8 ) = (x - 4) 2
2
√
(
x
9) = 42
x + 8 = x 2 - 8x + 16
x - 9 = 16
0 = x 2 - 9x + 8
x = 25
0 = (x - 1)(x - 8)
x - 1 = 0 or x - 8 = 0
x = 8 (x - 4 ≥ 0) ✓
21d
s = √
64 = √
21d
2
64 2 = ( √
21d )
4096 = 21d
195 ≈ d
200 - 195 = 5
There will be about 5 ft distance.
_1
PRACTICE AND PROBLEM SOLVING
2
( √
x - 12 ) = 9 2
2x + 1 = 27
2x = 26
x = 13
2x + 6 = 16
2x = 10
x=5
32. -3 = 2 √
x-7-7
√
2= x-7
x - 7)
2 = ( √
4=x-7
11 = x
2
2
33. √
4x + 12 = √
6x
= ( √
6x )
4x + 12 = 6x
12 = 2x
6=x
2
4x = x + 7
3x = 7
7
x=_
3
2
3x + 28 = x
0 = x 2 - 3x - 28
0 = (x - 7)(x + 4)
x - 7 = 0 or x + 4 = 0
x = 7 (x ≥ 0) ✓
43. √
x-3≤4
2
( √
x - 3) ≤ 42
2
x - 3 ≤ 16
x ≤ 19;
x-3≥0
x ≥ 3;
3 ≤ x ≤ 19
2
( √
8x + 1 ) ≥ 7 2
8x + 1 ≥ 49
8x ≥ 48
x ≥ 6;
8x + 1 ≥ 0
8x ≥ -1
1;
x ≥ -_
8
x≥6
25(x - 1) = x + 1
24x = 26
13
x=_
12
36. x + 3 = √
x+5
45.
2
x + 5)
(x + 3) = ( √
3
2
( √
3x + 28 ) = x 2
44. √
8x + 1 ≥ 7
2
2
(5 √
x - 1 ) = ( √
x + 1)
3
3
( √
4x ) = ( √
x + 7)
3
2
34. 5 √
x - 1 = √
x+1
2
2
≤6
3x + 3 ≤ 36
3x ≤ 33
x ≤ 11;
3x + 3 ≥ 0
3x ≥ -3
x ≥ -1;
-1 ≤ x ≤ 11
4
1 √
31. 3 = _
3x + 30
4
3x + 30
12 = √
3
( √
3x + 3 )
4
( √
2x + 6 ) = 2 4
x + 7 = 25
x = 18
3
4x = √
x+7
35. √
42. √
3x + 3 ≤ 6
4
30. √
2x + 6 = 2
2
( √
x + 7) = 52
( √
4x + 12 )
5x + 1 = 256
5x = 255
x = 51
3
29. 5 √
x + 7 = 25
√
x+7=5
2
4
3
( √
2x + 1 ) = 3 3
x - 12 = 81
x = 93
3x + 30 )
12 = ( √
144 = 3x + 30
114 = 3x
38 = x
4
( √
5x + 1 ) = 4 4
3
28. √
2x + 1 - 3 = 0
3
√
2x + 1 = 3
27. √
x - 12 = 9
_1
41. (3x + 28) 2 = x
√
3x + 28 = x
40. (5x + 1) 4 = 4
4
√
5x + 1 = 4
2
2
x + 6x + 9 = x + 5
x 2 + 5x + 4 = 0
(x + 1)(x + 4) = 0
x + 1 = 0 or x + 4 = 0
x = -1(x + 3 ≥ 0) ✓
327
√
15w
d=_
π
√
15w
1.5 = _
π
15w
1.5π = √
2
15w )
(1.5π) = ( √
22.2 ≈ 15w
1.5 ≈ x
About 1.5 tons weight can be lifted.
2
Holt McDougal Algebra 2
2
√
+ w2 + h2
13 2 + 5 2 + h 2
18 = √
2
18 2 = ( √194
+ h2)
d=
46a.
L
_
√
9.8
L
2.2 = 2π √_
9.8
T = 2π
52a.
2
0.35 ≈
√_πA
A
= ( √_
π)
r
2
48. r =
2
r3
A
r2 = _
π
3
3
3
b.
3V
r3 = _
4π
3V = 4πr 3
4 πr 3
V=_
3
A = πr 2
2E
√_
m
2E
= ( √_
m)
3V
√_
4π
3V
= ( √_
4π )
2
113 = k(2 +
√
b. No; if A = 80, then r must be less than or equal
to about 5.04. Therefore, 20 would not be a
reasonable value of r.
_3
2) 2
_3
113 = k(4) 2
113 = 8k
14 ≈ k
54. Solution B is incorrect; possible answer: in the first
step, the coefficient of the radical should have been
squared along with the rest of the equation.
_3
b. V = 14(6 + 2) 2 ≈ 317
The minimum wind velocity is about 317 mi/h.
_3
600 = 14(F + 2) 2
(F + 2) 3
42.86 ≈ √
(
3
55. x ≈ 5.84
57. x ≈ 2.35
)
b. d = 1.2116 √
h = 1.2116 √
120 ≈ 13.3
13.3 - 4.7 = 8.6
The sailor can see about 8.6 mi farther.
c. 8.6 mi ÷ 10 mi/h = 0.86 h ≈ 52 min
The sailor will see the pirate ship about 52 min
sooner than the captain will.
2
1837 ≈ (F + 2)
12 ≈ F + 2
10 ≈ F
The wind velocity is in the F10 category.
v = √ar
14 = √
39.2r
59.
2
2.5a )
8 = ( √
64 = 2.5a
25.6 = a
The acceleration is 25.6 m/s 2.
2
2
m
√_
ρ
m
= ( √_
ρ)
s=
s3
39.2r )
14 = ( √
196 = 39.2r
5=r
The smallest radius is 5 m.
b.
v = √
ar
8 = √
2.5a
2
56. x ≈ -2.38 or x ≈ -12.42
58a. d = 1.2116 √
h = 1.2116 √
15 ≈ 4.7
The captain can see about 4.7 mi.
42.86 ≈ √
(F + 2) 3
51a.
2
2
V = k(F + 2) 2
2
L
√_
9.8
60 = 2π _
L
_
√
120
9.8
L
0.08 ≈ √_
9.8
L
0.08 ≈ ( √_
9.8 )
T = 2π
53a. πr ≤ A
2
A
r ≤_
π
A
r≤ _
π
_3
c.
2
L
0.0064 ≈ _
9.8
0.06 ≈ L
The length of the pendulum is about 0.06 m.
2
2E
v2 = _
m
2E = mv 2
1 mv 2
E=_
2
50a.
(√ )
2
49. v =
v
9.8
L
_
9.8
L
0.1225 ≈ _
9.8
1.20 ≈ L
The length of the pendulum is about 1.20 m.
b. The length of the diagonal will double.
47. r =
L
_
0.35 ≈
2
324 = 194 + h
130 = h 2
11.4 ≈ h
The height of the prism is about 11.4 cm.
3
3
3
m
s3 = _
ρ
ρs 3 = m
m gold - m lead = 19.3(5) 3 - 11.34(5) 3 = 995
The mass of the cube of gold is 995 g greater.
60. Possible answer: Subtracting 5 from each
side results in the equation √
5x + 17 = -7.
Because the √ symbol indicates the principal, or
nonnegative, square root, the value of the left side
328
Holt McDougal Algebra 2
of the equation cannot be negative. Therefore, the
equation has no real solution.
73.
61. Possible answer: When solving both types of
equations, you must check for extraneous solutions.
√
x 2 - 64 = x - 4
2
( √
x 2 - 64 ) = (x - 4) 2
x 2 - 64 = x 2 - 8x + 16
8x = 80
x = 10
TEST PREP
62. D;
3
√
2x + 4 = 3
74. S 17 =
3
( √
2x + 4 ) = 3 3
2x + 4 = 27
2x = 23
x = 11.5
2.81S 4 - S 4
__
5x - 9 )
(x - 1) = ( √
x - 2x + 1 = 5x - 9
x 2 - 7x + 10 = 0
(x - 2)(x - 5) = 0
x - 2 = 0 or x - 5 = 0
x = 2 or x = 5
75a. D(n) = 2.00n + 5.00
2
2
b.
C
#OST
D
65. F;
40 = π √
82 + h2
64 + h 2
12.7 ≈ √
2
( √
64 + h 2 )
()
A
V= _
6
_3
2
√(_A6 )
A
= ( √(_
6) )
V=
2
V2
3
3
()
2
A
V = _
6
3
2
A
_
√
V =
6
0EOPLE
c. vertical translation 5 units down
2
76. f -1(x) = 2x - 8
3
78. f
-1
(x) = 7x + 2
1x - _
1
77. f -1(x) = -_
3
3
79.
_2
_1
66. (2x - 3) 4 = 3
4
√
2x - 3 = 3
4
( √
2x - 3 ) = 3 4
4
x·x
_
CHALLENGE AND EXTEND
68. never true
69. always true
70. never true
√
x
· √
x=9
( √x ) 2 = 9
x=9
√
x
8
4
√
x4 · x4
_
4
√4 3
4
4
√
x 4 · √
x4
_
2x - 3 = 81
2x = 84
x = 42
71.
x
√_
81
4
80.
67. always true
3
3
2
x
_
3
81.
√
3
18x 2
_
x4
18x
_
√x
3
3
√
18x
_
3
√
x3
3
√
18x
_
3
x
3
√
72. √
x+2 =4
2
√
( √
x + 2 ) = 42
READY TO GO ON? PAGE 637
√
x + 2 = 16
( √
x + 2)
√
64x 9
3
√
43 · x3 · x3 · x3
3
3
3
3
√
4 3 · √
x 3 · √
x 3 · √
x3
4·x·x·x
4x 3
6V 3 = A
9
√
x=_
4
SPIRAL REVIEW
2
64. B;
hm ≈ 2.81S
√_
36
× 100% = 181%
S4
The surface area increased by about 181%.
63. H;
x - 1 = √
5x - 9
162 ≈ 64 + h
98 ≈ h 2
10 ≈ h
7.875 ·
= √
3
12.7 2 ≈
(1 + 75%)h(1 + 350%)m
7.875hm
= √_
√___
36
36
2
1.
2
= 16
x + 2 = 256
x = 254
√
32x 3
√
2 2 · 2 2 · x 2 · 2x
√
2 2 · √
2 2 · √
x 2 · √
2x
2 · 2 · x · √
2x
√
4x 2x
329
Holt McDougal Algebra 2
2.
y 12z 6
√8
23 · y3 · y3 · y3 · y3 · z3 · z3
√
√
2 3 · √
y 3 · √
y 3 · √
y 3 · √
y 3 · √
z 3 · √
z3
3
17. -2 √
5x - 5 = -10 18. √x
+4=x-8
3
2
√
5x - 5 = 5
( √
x + 4 ) = (x - 8) 2
3
3
2
( √
5x - 5 ) = 5 3
x + 4 = x - 16x + 64
5x - 5 = 125
0 = x 2 - 17x + 60
5x = 130
0 = (x - 5)(x - 12)
x = 26
x - 5 = 0 or x - 12 = 0
3
3
3
3
3
3
3
3
3
2·y·y·y·y·z·z
2y 4z 2
√_a9
x = 12 (x - 8 ≥ 0) ✓
_3
4
4
3.
4. 4 2
√
43
√4 a4
_
√4 32
19. 3 √
x - 2 = √
6x
3
3
3
(3 √
x - 2 ) = ( √
6x )
3
√
64
8
√
32
a ·_
_
4
4
3 2 √
32
√
√
(√
32
a √
_
4
√4 34
3
a √
_
3
_5
_2
5. 16 4
6. (-27) 3
2
( √
-27 )
3
4
(2) 5
32
(-3) 2
9
_3
_2
7. 8 4
8. 243 5
168
_
n(168) = 112 · 2 50 ≈ 1150
The population is about 1150 after 1 week.
11. D: {x | x ≥ 0};
R: {y | y ≤ 4}
12. D: ;
R: y
y
64(x - 6) ;
√
STUDY GUIDE: REVIEW, PAGES 638-641
1. rational function
16.
y
y
2. direct variation; constant of variation
x
LESSON 8-1
1x
3. y = _
3
x
2x ≥ -8
x ≥ -4
14. g(x) = - √
x-2-3
64(10 - 6) = 16
g(10) = √
The speed is 16 ft/s.
3
x - 6 ≤ 196
x ≤ 202;
x-6≥0
x ≥ 6;
6 ≤ x ≤ 202
x
x
15.
3
( √
2x ) ≥ (-2) 3
2
( √
x - 6 ) ≤ 14 2
13. g(x) =
3
x + 5 < 16
x < 11;
x+5≥0
x ≥ -5;
-5 ≤ x ≤ 11
23. √
x - 6 - 10 ≤ 4
√
x - 6 ≤ 14
10. t = 24 × 7 = 168
≥ -2
22. √2x
2
( √
x + 5) < 42
_2
)
x+5<4
21. √
9. (-1000) 3
3
27(x - 2) = 6x
21x = 54
18
x=_
7
3 _
4w
20.
d=
0.02847
3 _
4w
7=
0.02847
3
3 _
3
4w
7 =
0.02847
4w
_
343 =
0.02847
9.77 ≈ 4w
2.4 ≈ w
The weight of the cultured pearl is about 2.4 carats.
4
5
( √
16 )
3
4. y = 4x
y
x
x
y
330
Holt McDougal Algebra 2
5.
n = ka
180 = k(20)
9=k
n = 9a = 9(34) = 306
The number of tiles needed is 306.
6.
I = kPr
264 = k(1100)(0.12)
2=k
I = 2Pr
360 = 2P(0.09)
360 = 0.18P
2000 = P
The principle P is $2000.
6
7. y = _
19.
x
x
k
I=_
R
k
8=_
15
120 = k
120
I=_
R
120
5=_
R
24 = R
The resistance R is 24 ohms.
22.
15.
x + 5x + 4
(x + 4)(x - 3)
__
(x + 4)(x + 1)
x - 3;
_
x+1
x ≠ -4 and x ≠ -1
-x + 2
x ·_
_
x-4
x2 + x - 6
-(x - 2)
x · __
_
x - 4 (x + 3)(x - 2)
-x
__
(x - 4)(x + 3)
x2 + 8
4
_
+_
2
3
6x
_
14.
9x + 3
x+5 _
_
·
3x + 1
x 2 - 25
3x + 1
(x + 5)(x - 5)
x-5
16.
2
x + 2x - 3 _
_
· x-2
x2 - x - 2
x+3
(x + 3)(x - 1) _
__
· x-2
(x - 2)(x + 1)
x-1
_
x+1
x+3
x+3
x +4
x +4
4 + x2 + 8
_
x-3
1 _
_
x2 + 4
2
x + 12
_
x-3+x+3
__
(
x+3 x-3
)
x-3
(
)
x+3
1 _
+_
x-3 x+3
(x + 3)(x - 3)
2x
__
; x ≠ ±3
(x + 3)(x - 3)
(
3(3x + 1)
x + 5 __
_
·
1 +_
1
_
x-2
x2 - 4
x+2
1 _
x
__
+_
x-2 x+2
(x + 2)(x - 2)
x+x+2
__
(x + 2)(x - 2)
2x + 2
__
(x + 2)(x - 2)
2(x + 1)
__
; x ≠ ±2
(x + 2)(x - 2)
3x + 12
3
6x
_
3(x + 4)
3
2x ; x ≠ -4
_
x+4
3
_
23.
2
1
x
+_
24. _
12.
14
2
x-2
x 2 + 2x - 8
(x
+
1)(x
+
3)
x-2
__ · _
(x + 4)(x - 2) 3(x + 1)
x+3
_
3(x + 4)
x2 + 4
LESSON 8-2
x 2 + x - 12
_
x 2 + 4x + 3 _
3x + 3
_
÷
LESSON 8-3
10. inverse variation
13.
(x + 3)(x - 3)
2
y
24x
11. _
9x 16
24 · x 14 - 16
_
9
_8 x -2
3
8 ;x≠0
_
3x 2
x-2
3x - 21 ÷ _
x - 49
20. _
3x
x 2 + 7x
x(x + 7)
3(x - 7) __
_
·
3x
(x - 7)(x + 7)
x-7 ·_
x
_
x
x-7
1
21.
9.
2x - 4
x+3
x
x-2
2(x + 5)
_
y
2
2
x + 2x - 15 _
__
÷ x -9
(x + 5)(x - 3) __
2(x - 2)
__
·
4
8. y = _
x
3
2
x y
x+3
9x - 1 · _
x
18. _ ÷ _
17. _
x 2 - 9 3x + 1
4xy 4
8y 2
(3x + 1)(3x - 1) _
x+3
2
__
8y 2
x ·_
_
·
3x + 1
(x + 3)(x - 3)
x
4y 3
3x - 1
_
2x
_
x-3
y
)
6
2x - 3 + _
25. _
3x + 7
4x - 1
3x + 7
2x - 3 _
6
4x - 1 + _
_
_
3x + 7 4x - 1
4x - 1 3x + 7
(2x - 3)(4x - 1) + 6(3x + 7)
___
(3x + 7)(4x - 1)
2
8x + 4x + 45
7 and x ≠ _
1
__
; x ≠ -_
3
4
(3x + 7)(4x - 1)
(
)
(
)
2
26. x - 9 = (x + 3)(x - 3)
x 2 - 6x + 9 = (x - 3) 2
The LCD is (x + 3)(x - 3) 2.
331
Holt McDougal Algebra 2
27. x 2 + 2x - 35 = (x + 7)(x - 5)
x 2 + 9x + 14 = (x + 7)(x + 2)
The LCD is (x - 5)(x + 2)(x + 7).
28.
3
2x - _
_
x+4
x+4
2x - 3 ; x ≠ -4
_
29.
35.
520
x+5
(
x-5
)
(
x+5
5 _
-_
x+5 x-5
x-5 x+5
x(x - 5) - 5(x + 5)
__
(x + 5)(x - 5)
2
x - 10x - 25 ; x ≠ ±5
__
(x + 5)(x - 5)
x+4
1
x
30. _
-_
31.
)
2x + 1
3x - 1
(
)
(
()
(
)
x
x
)
5(x + 2)
_x - _1
x
_
34. 4
x+2
_
x-2
_x 4x(x - 2) - _1 4x(x - 2)
x
4
___
x+2
_
4x(x - 2)
x-2
2
x (x - 2) - 4(x - 2)
__
4x(x + 2)
y
6x - 9
2
x+3 _
_
÷ x -9
3x
6x - 9
3(2x - 3)
x + 3 __
_
·
3x
(x + 3)(x - 3)
2x - 3
_
x(x - 3)
8(x - 6)
_
38. asymptotes: x = 1, y = -3;
D: {x | x ≠ 1}; R: {y | y ≠ -3}
39. asymptotes: x = -2, y = 1;
D: {x | x ≠ -2}; R: {y | y ≠ 1}
41. zero: 3;
asymptotes: x = -5,
x = -1, y = 0
40. zeros: 0, 3;
asymptote: x = -4
y
x+3
_
3x
_
33.
2
x -9
_
x+2
5
LESSON 8-4
y
2x + 1
7
_
-_
2x + 1 3x - 1
3x - 1 2x + 1
2x(3x - 1) - 7(2x + 1)
__
(3x - 1)(2x + 1)
2
6x - 16x - 7 ; x ≠ -_
1 and x ≠ _
1
__
2
3
(2x + 1)(3x - 1)
x-6
_
5
32. _
x+2
_
8
x+2
x-6 ÷_
_
5
8
x-6 ·_
8
_
55d
The jet’s average speed is about 548 mi/h.
7
2x - _
_
3x - 1
2x _
_
)
36. g is f translated 4 units 37. g is f translated 2 units
right.
right and 3 units up.
x+2
x2 - x - 6
x-3
1
x _
__
-_
x+2 x-3
(x - 3)(x + 2)
1 - x(x - 3)
__
(x - 3)(x + 2)
2
-x + 3x + 1
__
; x ≠ -2 and x ≠ 3
(x - 3)(x + 2)
(
580
2d(15080)
___
d (15080) + _
d (15080)
_
520
580
30160d
_
29d + 26d
30160d ≈ 548
_
5
x -_
_
x-5
x _
_
2d
_
d +_
d
_
y
x
x
43. zeros: -3, 3;
asymptote: x = 2
42. zero: 2;
asymptotes: x = -3,
y=2
y
y
x
()
x
44. hole at x = -3
(x - 4)(x - 2)
__
y
2
x
4x(x + 2)
(x + 2)(x - 2)(x - 2)
__
(OLEAT
x
4x(x + 2)
2
(x - 2)
_
4x
332
Holt McDougal Algebra 2
LESSON 8-5
LESSON 8-6
6 =1
45. x - _
x
6 (x) = 1(x)
x(x) - _
46.
47.
48.
x
2
x -6=x
x2 - x - 6 = 0
(x - 3)(x + 2) = 0
x = 3 or x = -2
3x + 5
4x = _
_
x-5
x-5
3x + 5
4x (x - 5) = _
_
(x - 5)
x-5
x-5
4x = 3x + 5
x=5
The solution x = 5 is extraneous.
Therefore there is no solution.
2x + 2
3x = _
_
x+2
x+2
2x + 2
3x (x + 2) = _
_
(x + 2)
x+2
x+2
3x = 2x + 2
x=2
2x
x =_
x +_
_
2
x+4
2x + 8
x 2(x + 4) + _
x 2(x + 4) = _
2x
_
2(x + 4)
2
x+4
2(x + 4)
2x + x(x + 4) = 2x
x(x + 4) = 0
x = 0 or x = -4
The solution x = -4 is extraneous.
The only solution is x = 0.
x+4
_
> -2
x
x is negative.
x is positive.
x+4
x+4
_
_
x > -2(x)
x < -2(x)
x
x
x + 4 > -2x
x + 4 < -2x
3x > -4
3x < -4
4
4
_
x>x < -_
3
3
The solution in this case
The solution in this case
4.
is x > 0.
is x < -_
3
4 or x > 0.
The solution to the inequality is x < -_
3
(
49.
(
50.
51.
)
(
()
)
(
)
)
3
52.
3x 2
_2
3
54. (-27) 3
√
23 · x3
_
3
3
√
3
3
√
2 ·
√
x3
_
3
2x
_
3
3
√
3
3
√
3
3
3
√
√3
3 _
2x · _
_
· 3
3
3
√
√
√
3
3
3
3
√
2x
9
_
3
_3
_3
55. 16 4
56. 9 2
_1
_1
_2
58. (9 4) 2
57. 17 3 · 17 3
1
4·_
_1 + _2
9 2
9 2 = 81
17 3 3
17 1 =_1 17
1 4
59. _
16
4 1
_
16
4
√
1
_
4
√
16
_1
2
( )
√
LESSON 8-7
60. D: {x | x ≥ 0};
R: {y | y ≥ 5}
61. D: ;
R: y
y
x
x - 3 is negative.
2 (x - 3) > 4(x - 3)
_
x-3
2 > 4x - 12
14 > 4x
7;
x<_
2
x-3<0
x<3
The solution in this case
is x < 3.
√4 81x 12
√4 34 · x4 · x4 · x4
4
4
4
4
4
4
4
√4 3 · √
x · √
x · √
x
3x 3
8x
√_
3
3
53.
2 <4
_
x-3
x - 3 is positive.
2 (x - 3) < 4(x - 3)
_
x-3
2 < 4x - 12
14 < 4x
7;
x>_
2
x-3>0
x>3
The solution in this case
7.
is x > _
2
√
27x 6
3
√
33 · x3 · x3
3
3
3
3
3
3
√
3 · √
x · √
x
x
333
y
63. h is f compressed
horizontally by a factor
1.
of _
4
y
x
x
7.
The solution to the inequality is x < 3 or x > _
2
62. g is f reflected across
the x-axis and
translated 1 unit up.
Holt McDougal Algebra 2
_1
64. j is f reflected across
65. k is f reflected across
the y-axis and
the x-axis, compressed
translated 8 units right.
vertically by a factor
1 , and translated
y
of _
2
1 unit up.
3
( √
x + 3 ) = (-6) 3
3
x + 3 = -216
x = -219
y
x
78. (x + 3) 3 = -6
3
√
x + 3 = -6
x
80. √
2x + 7 - 6 > -1
√
2x + 7 > 5
2
( √
2x + 7 ) > 5 2
66. g(x) = 3 √
x+4
67.
68.
y
2x + 7 > 25
x > 9;
2x + 7 ≥ 0
2x ≥ -7
x ≥ -3.5;
x>9
y
x
x
79. √x
-4≤3
2
( √
x - 4) ≤ 32
x-4≤9
x ≤ 13;
x-4≥0
x ≥ 4;
4 ≤ x ≤ 13
81. √
3x - 4 < 2
√
3x < 6
2
( √
3x ) < 6 2
3x < 36
x < 12;
3x ≥ 0
x ≥ 0;
0 ≤ x < 12
82. √x
- 1 > -2
3
3
( √
x - 1 ) > (-2) 3
3
LESSON 8-8
x - 1 > -8
x > -7
L
83.
T = 2π _
9.8
L
2.5 = 2π _
9.8
L
0.4 ≈ _
9.8
2
2
L
_
0.4 ≈
9.8
L
0.16 ≈ _
9.8
1.6 ≈ L
The length of the pendulum is about 1.6 m.
√
2x - 2
70. _ = 1
3
69. √
x + 6 - 7 = -2
√x
+6=5
6
3
√
2x - 2 = 6
2
( √
x + 6) = 52
3
( √
2x - 2 ) = 6 3
3
x + 6 = 25
x = 19
71. √
10x = 3 √
x+1
2x - 2 = 216
x = 109
√
(√
5
72. 2 √
x = √
64
5
2
2
( √
10x ) = (3 √
x + 1)
5
5
x ) = ( √
64 )
2( √
32x = 64
x=2
5
10x = 9(x + 1)
10x = 9x + 9
x=9
5
73. √
6x - 12 = x - 2
84.
6x - 12 = x 2 - 4x + 4
x 2 - 10x + 16 = 0
(x - 2)(x - 8) = 0
x = 2 or x = 8
75. (4x + 7) 2 = 3
√
4x + 7 = 3
2
3
512 = 6V √2
60.3 ≈ V
The volume of the tetrahedron is about 60.3 m 3.
4x + 7 = 9
x = 0.5
_1
76. (x - 4) 4 = 3
4
√
x-4=3
77. x = (2x + 35) 2
x = √
2x + 35
2
2x + 35 )
x = ( √
4
( √
x - 4) = 34
4
x - 4 = 81
x = 85
3
)
( √
6V √2
2
( √
4x + 7 ) = 3 2
2
x + 1 = x - 10x + 25
2
x - 11x + 24 = 0
(x - 3)(x - 8) = 0
x = 8 (x - 5 ≥ 0) ✓
_1
_1
3
8 =
_1
= (x - 5)
2) 3
s = (6V √
8 = (6V √
2) 3
74. √
x+1=x-5
( √
x + 1)
)
_1
2
( √
6x - 12 ) = (x - 2) 2
2
√
√
2
2
x = 2x + 35
x - 2x - 35 = 0
(x - 7)(x + 5) = 0
x = 7 (x ≥ 0) ✓
2
334
Holt McDougal Algebra 2
8. Let n be the number of words on three pages.
2n
_
n +_
n
_
62
45
2n(2790)
__
n (2790) + _
n (2790)
_
62
45
5580n
_
45n + 62n
5580n ≈ 52
_
107n
Her average typing speed is about 52 words/min.
CHAPTER TEST, PAGE 642
1.
2.
p = kb
19.80 = k(1100)
k = 0.018
p = 0.018b = 0.018(3000) = 54
The payment p is $54.
k
t=_
v
25 = _
k
_
60
6
k = 2.5
2.5
t =_
v
20 = _
2.5
_
v
60
v = 7.5
The speed v has to be 7.5 mi/h.
9. zero: -1;
asymptotes: x = -2, y = 3
x
x-5
x-9 ·_
x -x-6
4. _
3. _
2
2x
10
x - 4x + 3
x 2 - 81
x-5
x
9
_
(x
3)(x
+
2)
__
· __
2(x - 5) (x - 9)(x + 9)
(x - 3)(x - 1)
1
_
x+2
_
; x ≠ 1 and x ≠ 3
2(x
+ 9)
x-1
2
5 +_
3x - 9x ÷ __
2x - 6
x
6. _
5. _
2
2
x
5
2x
10
x - 16
x - 8x + 16
5 _
2 +_
x
_
2
3
2
x - 8x + 16
3x - 9x · __
_
x
5
2
2(x
- 5)
2x - 6
x 2 - 16
5(2)
+
x
_
2
2
(x - 4)
3x (x - 3)
__
2(x - 5)
·_
(x + 4)(x - 4) 2(x - 3)
x + 10
_
;x≠5
2
2(x
- 5)
(x
4)
3
x
_
2(x + 4)
9x - 6
_
7. 5x - _
x-7
x+3
x+3
5x _
9x - 6 _
x-7
_
-_
x-7 x+3
x+3 x-7
5x(x + 3) - (9x - 6)(x - 7)
___
(x - 7)(x + 3)
3
2
()
(
)
(
)
2
2
5x + 15x - (9x - 69x + 42)
___
(x - 7)(x + 3)
2
-4x + 84x - 42
__
(x - 7)( x+ 3)
y
2
-2(2x - 42x + 21)
__
; x ≠ 7 and x ≠ -3
(x - 7)(x + 3)
3 = 10
10. 2 + _
x-1
3 =8
_
x-1
3 (x - 1) = 8(x - 1)
_
x-1
3 = 8x - 8
11 = 8x
11 = x
_
8
5
x
x
_
_
+ =_
11.
3
x-1
x-1
5 3(x - 1)
x 3(x - 1) + _
x 3(x - 1) = _
_
3
x-1
x-1
3x + x(x - 1) = 15
x 2 + 2x - 15 = 0
(x - 3)(x + 5) = 0
x = 3 or x = -5
1 (2.4) + _
1 (2.4) = 1
12. _
6
h
_1 (2.4)(6h) + _1 (2.4)(6h) = 1(6h)
6
h
2.4 h + 14.4 = 6h
14.4 = 3.6h
4=h
It would take Mike 4 hours to tile the floor alone.
335
(
)
()
(
)
Holt McDougal Algebra 2
13.
√
-32x 6
3
14. 8
1
_
)3 · 4 · x3 · x3
√(-2
(-2) 3 · √x3 · √
x 3 · √4
√
3
3
3
3
2
-_
3
√
82
3
3
1
_
2
)
-2x ( √4
3
3
√
64
_1
4
_2
_2
16. x 5
27 3
15. _
_1
27 3
_2 - _1
27 3
3
_1
27 3
3
√27
= 3
17. D: {x | x ≥ -2};
R: {y | y ≥ -4}
18.
y
y
x
x
19. √
x+7=5
2
( √
x + 7) = 52
x + 7 = 25
x = 18
_1
21. (3x + 1) 3 = -2
3
√
3x + 1
= -2
3
( √
3x + 1 ) = (-2) 3
3
3x + 1 = -8
3x = -9
x = -3
20. √
2x + 1 = √
x+ 9
2
2
( √
2x + 1 ) = ( √
x + 9)
2x + 1 = x + 9
x=8
22. s =
A
√_
4.828
12.4 =
A
_
4.828
2
A
_
12.4 =
4.828
2
A
_
12.4 =
4.828
742 ≈ A
The area is about 742 in 2.
2
(√
)
+1>3
23. √2x
2
( √
2x + 1 ) > 3 2
2x + 1 > 9
2x > 8
x > 4;
2x + 1 ≥ 0
2x ≥ -1
x ≥ -0.5;
x>4
336
Holt McDougal Algebra 2