Chapter 10 Summary and Review: Study Guide 281
104. 6x=12
106. (4.1×102)(6.5×106)=26.65 ×104
50
54i=50
54i·5+4i
5+4i=250 + 200i
41 =250
41 +200
41 i
110. 121
1·1
112. i5+i6+i7+i8
114. 55i
5i=55i
5i·5i
5i=55i5
5=
Chapter 10 Vocabulary Reinforcement
a, if the
4. To find an equivalent expression without a radical in the
6. An imaginary number is a number that can be named bi,
where bis some real number and b=0.
Chapter 10 Concept Reinforcement
1. True; see page 695 in the text.
3. False; radicals cannot be added unless they have the same
index and the same radicand.
Chapter 10 Study Guide
1. 36y2=(6y)2=|6y|=6|y|
25y26=5y6
8. 20a
5=20a
5=4a=2
a
11. 3+x1=x
x2=0 or x 5=0
12. x+3x2=1
x+3=x2+1
2=(
x2+1)
2
13. (2 5i)2=420i+25i2=420i+ 25(1) =
Chapter 10 Review Exercises
f(1) = 3·116 = 316 = 13
4. The domain of f(x)=3x16 is the set of all x-values
for which 3x16 0.
x16
3
The domain is x
x16
3,or16
3,.
10. 3
27 =1
3Since 1
33
=1
27
17. 5
a2b3=(a2b3)1/5
18. 491/2=1
491/2=1
49 =1
7
Chapter 10 Summary and Review: Review Exercises 283
1/3(1/2) =7
1/3+1/2=7
2/6+3/6=7
1/6
27. 4
x3
x=x1/4·x1/3=x1/4+1/3=x3/12+4/12 =x7/12 =
108 = 3
27 ·4= 3
27 3
4=33
4
31. 49
36 =49
36 =7
6
64x6
3
64x6
12
x5
40. 53
x+23
x=(5+2)3
x=73
x
44. (538)(5+2
8)
45. (1 7)2=1
22·1·7+(
7)2
46. (3
27 3
2)( 3
27 + 3
2)
47. 8
9=4·6
50. 4
x+3=2
51. x+83x+1=1
x+8=3x+1+1
y
284 Chapter 10: Radical Expressions, Equations, and Functions
52. d(n)=0.752.8n
(108)2=(
2.8n)2
54. Let s= the length of a side of the square, in cm. We use
the Pythagorean theorem.
55. Let w= the width of the bookcase, in ft. Then we can
refer to the drawing in the text, replacing “?” with w.We
use the Pythagorean theorem.
56. a=7,b=24
625 = c2
57. a=2,c=5
2
Find b.
a2+b2=c2
58. 25 + 8=1·25 + 1·4·2=
59. (4+3i)+(212i)=(4+2)+(312)i=
61. (2+5i)(2 5i)=2
2(5i)2
=1212i3i2=1
=912i
25
=10
25 +15
25i
=37i2
9+1
66. Graph: f(x)=x.
curve.
x f(x) (x, f(x))
0 0 (0,0)
Chapter 10 Test 285
67. We substitute each complex number in the equation to
determine if it is a solution.
12i1+44i+5
96iFALSE
For 1 + i:
x2+4x+5=0
For 2 + i:
x2+4x+5=0
Choice D is correct.
68. i·i2·i3…i
99 ·i100
Pairing iand i100,i2and i99,i3and i98, and so on, we
6+x=3611x
(6+x)2= (36 11x)2
121x430=0 or x 3=0
121x= 430 or x =3
Chapter 10 Discussion and Writing Exercises
Consider (x+5)
1/2. Since the exponent is 1
2,x+5
2. Since xexists only for {x|x0}, this is the domain of
y=x·x.
Chapter 10 Test
1. 148 12.166
3. The domain of f(x)=84xis the set of all x-values
for which 8 4x0.
286 Chapter 10: Radical Expressions, Equations, and Functions
5. x2+10x+25=(x+5)
2=|x+5|
9. a2/3=3
10. 323/5=5
323=5
(25)3=5
215 =2
3=8
17. 8
x2=x2/8=x1/4=4
x
16x6= (16x6)1/4=(2
3
16x5
3
16x5
3
8x3·2x2
26. 3
2x3
5y2=3
2x·5y2=3
10xy2
30. 3128+2
18+2
32
=3
64 ·2+2
9·2+2
16 ·2
31. (20+2
5)(20 35)
33. 1+2
352=1+2
352·3+5
2
3+5
2
35. x6=x+93
(x6)2=(
x+93)2
Chapter 10 Test 287
36. x1+3=x
37. We make a drawing, letting s= the length of a side of the
square, in feet.
72s
38. D=1.2h
72=1.2h
39. a=7,b= 7; find c.
a2+b2=c2Pythagorean equation
40. a=1,c=5; find b.
43. (3 4i)(3+7i)=9+21i12i28i2FOIL
=9+21i12i+28 i2=1
45. 7+14i
68i=7+14i
68i·6+8i
6+8i
x2+2x+5=0
(1+2i)2+2(1+2i)+5 ? 0
47. x4=x2
(x4)2=(
x2)2
288 Chapter 10: Radical Expressions, Equations, and Functions
48. 14i
4i(1+4i)1=(1 4i)(1+4i)
4i
=68i
16
49. 2x2+7x+4=13x+10
(2x2+7x+4)
2=(
13x+ 10)2
10x222x24 = 0
5x211x12 = 0 Dividing by 2
(5x+ 4)(x3) = 0
1. (2x23x+1)+(6x3x3+7x24) =
3x3+9x2+3x3
4. x3+64
x249 ·x214x+49
x24x+16
5.
y27y18
y2+3y+2
=(y25y6)(y2+4y+4)
(y27y18)(y2+3y+2)
6.
x
x+2+1
x3x22
x2x6
=x
x+2+1
x3x22
(x+ 2)(x3),
7. y2+y2
y+2 y3+3y2+0y5
y3+2y2
Cumulative Review Chapters 1 – 10 289
10. 975+6
12=9
25 ·3+6
4·3=
12. 35
63=35
63·6+3
6+3=330+3
15
13. 6
m12n24
64 =m12n24
261/6
=m2n4
2
17. 1
5+3
10x=4
5
x=10
3·3
5
19. 3a4<10+5a
2a4<10
20. 8<x+2<15
21. |3x6|=2
3x6=2or 3x6=2
3x=4 or 3x=8
22. 625 = 49y2
7y=25 or 7y=25
23. 3x+5y=30,(1)
5x+3y= 34 (2)
5y=15
y=3
24. 3x+2yz=7,(1)
x+y+2z=9,(2)
3x+2yz=7
5x+5y+z=1
35x+35y=35
x=1
Next we use Equation (4) to find y.
5(1) + 5y=5 Substituting
290 Chapter 10: Radical Expressions, Equations, and Functions
5(1)+5·0+z=1
5+z=1
6x
x5300
x2+5x+25 =2250
(x5)(x2+5x+25)
6x3+30x2+ 150x300x+ 1500 = 2250
6x3+30x2150x+ 1500 = 2250
x=5or x =5 or x =5
The number 5 checks but 5 does not, so the solution is
5.
26. 3x2
(x+ 2)(x2) ·48
(x+ 2)(x2)
27. I=nE
R+nr
28. 4x+12=3
4x+1=5
29. 21x=5
(21x)2=(
5)2
30. 13 x=5+x+4
8x=x+4
(8 x)2=(
x+4)
2
64 16x+x2=x+4
31. Graph: f(x)=2
3x+2
y
y
x
y
Cumulative Review Chapters 1 – 10 291
32. Graph: 4x2y=8
First we will find the intercepts. To find the x-intercept,
The x-intercept is (2,0).
We use a third point as a check. Let x=4.
33. Graph: 4x5y+20
34. Graph: y≥−3,
y2x+3
y=3,
35. Graph: g(x)=x2x2
Make a list of function values in a table.
g(0) = 0202=2
1 0
02
x
y
y
292 Chapter 10: Radical Expressions, Equations, and Functions
36. f(x)=|x+4|
Make a list of function values in a table.
x f(x)
5 1
37. Graph: g(x)= 4
51
2
38. Graph: f(x)=2x
We find some ordered pairs, plot points, and draw the
curve.
50.2 (5,0.2)
40. 3x217x28
We will use the ac-method.
2) Multiply the coefficients of the first and last terms.
41. y2y132
42. 27y3+8=(3y)3+2
3=(3y+ 2)(9y26y+4)
Cumulative Review Chapters 1 – 10 293
The smallest y-value is 5. No endpoints are indicated,
46. First we find the slope-intercept form of the equation by
47. First we find the slope of the given line.
48. h=k
b
100 = k
20
A=1
49. Familiarize. Let t= the number of hours it will take the
t
Check. We verify the work principle.
50. V=kd2
52. Using the work principle, we translate to the equation
t
3+t
15 =1,
53. 3112 4
294 Chapter 10: Radical Expressions, Equations, and Functions
54. 2x+6=8+5x+1
2x2=5x+1
55. x+x+1
xx+1 =5
11
9x+8=0 or x 8=0