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Document 1803765
```Chapter 4,
continued
}
7.
x 5 24
}
(n 2 8)(n 2 3) 5 0
}
n2850
or
n2350
n58
or
n53
}
z 2 3z 2 40 5 0
2
(z 1 5)(z 2 8) 5 0
10.
}
12.
12.
Ï
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}
}
}
}
}
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a54
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Ï51
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}
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221 2 3Ï5
}
}
}
(50 1 5x)
+
(140 2 10x)
R(x) 5 250(x 2 14)(x 1 10)
}
}
32 1 4Ï3
5}
61
17. 5x 2 5 80
x 5 16
18. z 2 2 7 5 29
z 2 5 36
2
}
}
x 5 6Ï16
z 5 6Ï 36
x 5 64
z 5 66
19. 3(x 2 2) 5 40
2
40
(x 2 2)2 5 }
3
14 1 (210)
2
x5}5}52
Î40
}
To maximize revenue, each DVD player should cost
140 2 10(2) 5 \$120.
x2256 }
3
R(2) 5 250(2 2 14)(2 1 10) 5 7200
x2256}
} + }
}
The maximum monthly revenue is \$7200.
}
}
Ï40
Ï3
Ï3
Ï3
}
Ï120
x2256}
3
Lesson 4.5
}
4.5 Guided Practice (pp. 266–269)
}
}
1. Ï 27 5 Ï 9 + Ï 3 5 3Ï 3
}
}
}
}
}
}
3. Ï 10 + Ï 15 5 Ï 150 5 Ï 25 + Ï 6 5 5Ï 6
}
}
}
Ï 4 + Ï 30
x2256}
3
2Ï30
}
2. Ï 98 5 Ï 49 + Ï 2 5 7Ï 2
}
}
29 1 Ï 7
R(x) 5 210 + 5(x 2 14)(x 1 10)
}
}
5}
74
R(x) 5 210(x 2 14)(5x 1 50)
}
}
}
}
R(x) 5 (210x 1 140)(5x 1 50)
}
}
32 1 4Ï 3
4
8 1 Ï3
4
16. }
} + }
}
}
} 5 }
} 5 }}
8 2 Ï3 8 1 Ï3
64 1 8Ï3 2 8Ï 3 2 3
8 2 Ï3
5
}
}
29 1 Ï7
21
9 2 Ï7
21
15. }
} + }
}
}
} 5 }
} 5 }}
9 1 Ï7 9 2 Ï7
81 2 9Ï7 1 9Ï7 2 7
9 1 Ï7
Price
Monthly
Sales
revenue 5 (dollars/DVD + (DVD players)
player)
(dollars)
}
}
}
}
5
}
}
}
8 2 2Ï11
x 5 2}2
}
}
5}
5
2x 1 5 5 0
}
}
4x 1 20x 1 25 5 0
p1q
2
}
8 2 2Ï 11
2
2
4 2 Ï 11
14. }
} + }
}
}
} 5 }
} 5 }}
4 1 Ï 11 4 2 Ï11
16 2 4Ï11 1 4Ï 11 2 11
4 1 Ï11
2
R(x)
}
}
}
}
(2x 1 5)(2x 1 5) 5 0
13.
}
242 2 6Ï5
or a 2 4 5 0
or
}
5}
5}
44
22
(7a 2 2)(a 2 4) 5 0
2
}
Ï19
Ï19 Ï21
Ï399
19
}5}
} 5 }
} + }
} 5 }
21
21
Ï 21
Ï 21 Ï 21
s53
or
7a 2 2 30a 1 8 5 0
a 5 }7
}
Ï15
26
242 2 6Ï 5
26
7 1 Ï5
13. }
} + }
}
}
} 5 }
} 5 }}
7 2 Ï5 7 1 Ï5
49 1 7Ï5 2 7Ï 5 2 5
7 2 Ï5
or s 2 3 5 0
7a 2 2 5 0
Ï15
}
(5s 1 1)(s 2 3) 5 0
11.
}
15
}5}
} 5 }
2
4
Ï4
}
}
}
5s2 2 14s 2 3 5 0
1
s 5 2}5
}
2Ï51
z58
5s 1 1 5 0
}
6.
5}
5}
12
6
or z 2 8 5 0
z 5 25 or
}
3
8
}
8. n 2 2 11n 1 24 5 0
z1550
Ï9
Ï64
}
x1450
or
x 5 25 or
9.
9
64
}5}
} 5 }
5.
(x 1 5)(x 1 4) 5 0
x1550
}
Ï
Ï
Ï36
Ï 11
Ï 11
6
36
11
7.
5}
8. } 5 } 5 }
5
Ï49 Ï49 7
Ï}25 5 }
Ï25
Ï6
Ï6 Ï5
Ï30
6
9. } 5 } 5 } + } 5 }
Ï 5 Ï5 Ï5 Ï5 5
Ï9
Ï9 Ï8
Ï72
Ï36 + Ï2
6Ï 2
3Ï 2
9
10. } 5 } 5 } + } 5 } 5 } 5 } 5 }
Ï 8 Ï8 Ï8 Ï8 8
8
8
4
Ï 4 + Ï 51
Ï17
Ï17 Ï12
Ï204
17
11. } 5 } 5 } + } 5 } 5 }
Ï12 Ï12 Ï12 Ï12 12
12
x 2 1 9x 1 20 5 0
}
}
4. Ï 8 + Ï 28 5 Ï 224 5 Ï 16 + Ï 14 5 4Ï 14
x2256}
3
}
2Ï30
x526}
3
Algebra 2
Worked-Out Solution Key
197
Chapter 4,
continued
}
}
0 5 216t 2 1 30
}
30
16
}
Î16
30
6 }
5t
}
}
}
}
}
}
}
}
61.4 ø t
Ï96 5 Ï4 + Ï24 5 2Ï24 5 2Ï4 + Ï6 5 4Ï6
Reject the negative solution, 21.4, because time must
be positive. The container will fall for about 1.4 seconds
before it hits the ground.
or Ï96 5 Ï 16 + Ï6 5 4Ï6
}
x569
22. s 2 5 169
2. To “rationalize the denominator” of a quotient containing
square roots means to eliminate the radical from the
denominator.
}
}
}
}
}
}
}
}
}
}
}
}
8. 5Ï 2 4 + 3Ï 10 5 15Ï 240 5 151 Ï 16 + Ï 15 2
}
}
Ï
}
}
Ï5
Ï5
5
}5}
} 5 }
4
16
Ï16
10.
Ï
}
}
}
}
}
}
Ï13
}
}
}
}
}
}
}
}
}
}
}
}
Ï2
4Ï2 2 Ï10
Ï2
4 2 Ï5
17. }
} + }
}
}
} 5 }
} 5 }}
4 1 Ï5 4 2 Ï5
16 2 4Ï5 1 4Ï5 2 5
4 1 Ï5
}
Algebra 2
Worked-Out Solution Key
x 5 610
31. 4(x 2 1)2 5 8
t2
20
}57
(x 2 1)2 5 2
}
x 2 1 5 6Ï2
t 2 5 140
}
}
}
}
5 2 Ï6
1
5 2 Ï6
1
16. }
} + }
}
}
} 5 }
} 5 }}
5 1 Ï6 5 2 Ï6
25 2 5Ï6 1 5Ï6 2 6
5 1 Ï6
}
5 2 Ï6
5}
19
}
x 5 6Ï100
}
t 5 6Ï 140
}
5 21 2 Ï 3
}
x 2 5 100
}
t2
30. } 1 8 5 15
20
2 1 2Ï3
2 1 2Ï 3
2
1 1 Ï3
2
15. }
5}
} + }
}
}
} 5 }
} 5 }}
22
1 2 Ï3 1 1 Ï3
1 1 Ï3 2 Ï3 2 3
1 2 Ï3
4Ï2 2 Ï 10
x2
25
}54
r 5 6Ï 5
}
Ï 4 + Ï 91
}
5}
11
x2
29. } 2 6 5 22
25
r2 5 5
}
}
}
w 5 6Ï71
}
7r 2 5 35
Ï91
2Ï91
5}
5}
28
14
}
w 2 5 71
28. 7r 2 2 10 5 25
}
Ï364
Ï28
}
27. 23w 2 5 2213
p 5 6Ï 16 + Ï 7
}
}
}
Ï13
z 5 65
}
5}+}5}
5}
Ï}28 5}
28
28
Ï28
Ï28 Ï28
13
x 5 62Ï21
}
}
}
z 5 6 Ï25
p 5 64Ï 7
}
}
}
x 5 6 Ï 4 + Ï21
p 5 6Ï 112
Ï18
Ï18 Ï11
Ï198
Ï9 + Ï22
3Ï22
Î}1811 5 }
5}+}5}
5}
5}
11
11
11
Ï 11
Ï 11 Ï 11
}
z 2 5 25
}
}
7
7Ï12
Ï 12
7
7Ï 3
12. }
} + }
} 5 }
} 5 } 5 }
12
6
Ï 12 Ï 12
Ï 12
}
}
p 5 112
}
Ï35
Ï35
35
}5}
} 5 }
6
36
Ï36
8
8Ï 3
Ï3
8
11. }
} + }
} 5 }
} 5 }
3
Ï3 Ï3
Ï3
}
25. 6z 2 5 150
2
}
}
}
}
26. 4p2 5 448
}
5 151 4Ï 15 2 5 60Ï 15
}
}
}
7. 4Ï 6 + Ï 6 5 4Ï 36 5 4(6) 5 24
}
a 5 6 Ï25 + Ï2
}
6. Ï 3 + Ï 27 5 Ï 81 5 9
}
s 5 6 13
x 5 6 Ï84
}
}
a 5 6Ï 50
24. x 2 5 84
5. Ï 150 5 Ï 25 + Ï 6 5 5Ï 6
}
}
s 5 6 Ï169
a 5 65 Ï2
4. Ï 192 5 Ï 64 + Ï 3 5 8Ï 3
}
23. a 2 5 50
}
x 5 1 6 Ï2
}
t 5 6Ï 4 + Ï35
}
t 5 62Ï 35
}
}
x 5 6 Ï 81
expression.
3. Ï 28 5 Ï 4 + Ï 7 5 2Ï 7
}
5x2 5 405
}
}
}
}
x 2 5 81
}
}
number solutions: x 5 Ï 81 , and x 5 2Ï81 .
1. In the expression Ï 72 , 72 is called the radicand of the
}
}
21. Because 81 > 0, the equation x 2 5 81 has two real-
Skill Practice
198
}
20. The expression was not completely simpliﬁed.
4.5 Exercises (pp. 269–271)
14.
}
19. C; Ï 108 5 Ï 36 + Ï 3 5 6Ï 3
}
13.
}
}
5 2}}
6
} 5 t2
}
}
}
6 1 3Ï 10 1 2Ï7 1 Ï70
230 5 216t 2
9.
}
}
3 1 Ï 7 2 1 Ï10
6 1 3Ï 10 1 2Ï7 1 Ï 70
3 1 Ï7
18. }
} + }
}
}
} 5 }
} 5 }}
2 2 Ï 10 2 1 Ï10
4 1 2Ï 10 2 2Ï 10 2 10
2 2 Ï10
h 5 216t 2 1 h0
20.
Chapter 4,
continued
32. 7(x 2 4)2 2 18 5 10
33. 2(x 1 2)2 2 5 5 8
Problem Solving
7(x 2 4) 5 28
2(x 1 2)2 5 13
(x 2 4)2 5 4
(x 1 2)2 5 }
2
2
h 5 216t2 1 40
13
Ï
}
}
}
Ï26
x 1 2 5 6}
2
x 5 4 6 2 5 6, 2
}
Ï26
x 5 22 6 }
2
34. C; 3(x 1 2)2 1 4 5 13
40
16
} 5 t2
Ï
}
40
6 }
5t
16
61.6 ø t
Reject the negative solution, 21.6, because time must be
positive. The diver is in the air for about 1.6 seconds.
g
39. h 5 2} t 2 1 h0
2
32 2
Earth: 0 5 2}
t 1 150
2
3(x 1 2) 5 9
2
}
x 1 2 5 6Ï3
240 5 216t 2
}
Ï13
Ï2
x 1 2 5 6}
} 2 }
}
Ï2
Ï2
x 2 4 5 62
0 5 216t2 1 40
}
13
x1256 }
2
x 2 4 5 6Ï4
(x 1 2)2 5 3
h 5 216t2 1 h0
38.
0 5 216t 2 1 150
}
x 5 22 6 Ï3
2150 5 216t 2
35. One method for solving the equation is to use the special
150
16
} 5 t2
factoring pattern known as the difference of two squares.
Ï
x2 2 4 5 0
(x 1 2)(x 2 2) 5 0
x1250
}
150
6 }
5t
16
or
x 5 22 or
x2250
x52
Another method for solving the equation is to use
square roots.
63.1 ø t
It takes the rock about 3.1 seconds to hit the surface
of Earth.
12
t 2 1 150
Mars: 0 5 2}
2
0 5 26t 2 1 150
x2 2 4 5 0
x2 5 4
2150 5 26t2
}
x 5 6 Ï4
25 5 t 2
x 5 62
}
6Ï 25 5 t 2
65 5 t
a. x 5 121
2
It takes the rock 5 seconds to hit the surface of Mars.
b. x 2 5 0
76
t 2 1 150
Jupiter: 0 5 2}
2
c. x 5 236
2
0 5 238t 2 1 150
37. a(x 1 b)2 5 c
2150 5 238t 2
c
(x 1 b)2 5 }a
150
38
} 5 t2
Îc
}
x 1 b 5 6 }a
Ï
}
150
Ïc
}
Ïa
6 }
5t
38
Ïa
Ïa
62 ø t
}
x 1 b 5 6}
} + }
}
}
Ï ca
x 1 b 5 6}
a
}
Ï ca
x 5 2b 6 }
a
It takes the rock about 2 seconds to hit the surface
of Jupiter.
30
t2 1 150
Saturn: 0 5 2}
2
0 5 215t 2 1 150
2150 5 215t 2
10 5 t 2
}
6Ï 10 5 t
63.2 ø t
It takes the rock about 3.2 seconds to hit the surface
of Saturn.
Algebra 2
Worked-Out Solution Key
199
Chapter 4,
continued
2
Pluto: 0 5 2}2 t 2 1 150
0 5 2t 2 1 150
05
150 5 t 2
}
2
2 }
2
Ï3
}t 2
1 Ïh0 2 2:d
*w
}
}
6Ï150 5 t
}
2:d 2Ï3
*w
0 5 Ï h0 2 } t
612.2 ø t
}
It takes the rock about 12.2 seconds to hit the surface
of Pluto.
h 5 0.019s2
5 5 0.019s2
20 5 0.019s2
} 5 s2
5
0.019
} 5 s2
}
}
5
20
0.019
Ï
20
6 }
5s
0.019
6 }
5s
0.019
616.2 ø s
632.4 ø s
16.2 knots
32.4 knots
The wind speed required to generate 20 foot waves is twice
the wind speed required to generate 5 foot waves.
41. a. Area of circle 5 Area of square
:r 2 5 10 2
:r 2 5 100
}
2:d 2Ï3
} t 5 Ï h0
*w
}
*wÏh0
}
}
*wÏ3h0
Ï3
t5}
} + }
} 5 }
2
6:d 2
2:d Ï3 Ï3
Mixed Review for TAKS
44. B;
The line is dashed, so choices C and D can be eliminated.
The half-plane above the line is shaded, so choice A can be
eliminated. The inequality y > 2x 2 3 is graphed.
45. H;
3x 1 y 5 21 l y 5 23x 2 1 l m 5 23
1
1
2x 1 3y 5 6 l y 5 }3 x 1 2 l m 5 }3
The lines are perpendicular because 231 }3 2 5 21.
1
Problem Solving Workshop 4.5 (p. 273)
b. :r 5 100
2
1. 2x 2 2 12x 1 10 5 0
100
r2 5 }
:
x51
}
Ï
100
r56 }
:
r ø 65.6
c. :r 2 5 s2
X
0
1
2
3
4
5
6
X=1
or
x55
Y1
10
0
-6
-8
-6
0
10
s2
r2 5 }
:F
2. x 2 1 7x 1 12 5 0
Î
}
s2
r56 }
:
x 5 24
R 5 0.00829s2
42. a.
5 5 0.00829s
Î
2
5
} 5 s2
0.00829
}
5
X
-6
-5
-4
-3
-2
-1
0
X=-4
or
x 5 23
Y1
6
2
0
0
2
6
12
6 }
5s
0.00829
624.6 ø s
Reject the negative solution, 224.6, because speed must
be positive. The speed of the racing cyclist is about
24.6 miles per hour.
b. R 5 0.00829s2
R 5 0.00829(2s)2
R 5 0.00829(4s2)
R 5 4(0.00829s2)
The air resistance quadruples when the cyclist’s speed
doubles.
200
Algebra 2
Worked-Out Solution Key
3. 9x 2 2 30x 1 25 5 0
x ø 1.7
X
1.4
1.5
1.6
1.7
1.8
1.9
2.0
X=1.7
Y1
.64
.25
.04
.01
.16
.49
1
o
r
e
Z
X=1.6666666 Y=0
Ï
}
The pool is completely drained when h 5 0.
2150 5 2t 2
40.
2 }
Ï3
} t2
1 Ïh0 2 2:d
*w
h5
43.
Chapter 4,
continued
4. 7x 2 2 3 5 0
X
-1
-.9
-.8
-.7
-.6
-.5
-.4
X=-.7
Y1
4
2.67
1.48
.43
-.48
-1.25
-1.88
8. h 5 216t 2 1 h0
X
.2
.3
.4
.5
.6
.7
.8
X=.6
0 5 216t2 1 30
Y1
-2.72
-2.37
-1.88
-1.25
-.48
.43
1.48
X
Y1
1
14
1.1
10.6
4
1.2
.9
66
1.3
2.9
6
1.4
1.36
1.5
6
1.6
10.9
- 6
X=1.4
Zero
X=1.36
306
94
The shellﬁsh hits the
ground between 1.3 and
1.4 seconds after it is
dropped.
Y=
0
The shellﬁsh hits the
seconds after it is
dropped.
9. h 5 216t 2 1 h0
The tables show that x is between 20.7 and 20.6 or x is
between 0.6 and 0.7.
5. x 1 3x 2 6 5 0
4 5 216t2 1 29
0 5 216t2 1 25
2
X
-4.8
-4.7
-4.6
-4.5
-4.4
-4.3
-4.2
X=-4.4
Y1
2.64
1.99
1.36
.75
.16
-.41
-.96
X
1
1.1
1.2
1.3
1.4
1.5
1.6
X=1.4
X
Y1
1
9
1.1
5.6
4
1.2
1.9
6
1.3
2.04
1.4
.36
6
1.5
11
1.6
15.9
- 6
X=1.3
Y1
-2
-1.49
-.96
-.41
.16
.7
5
1.36
Zero
X=1.25
The ball is in the air between
1.2 and 1.3 seconds before
Y=0
The ball is in the air 1.25
seconds before your
friend catches it.
10. h 5 16t 2 1 h0
0 5 216t2 1 50
The tables show that x is between 24.4 and 24.3 or x is
between 1.3 and 1.4.
6. h 5 216t 2 1 h0
The container hits the ground between 1.76 and
1.77 seconds after it is dropped.
0 5 216t 1 100
2
X
2
2.1
2.2
2.3
2.4
2.5
2.6
X=2.5
Y1
36
29.44
22.56
15.36
.84
7
0
-8.16
o
r
e
Z
X=2.5
X
1.73
1.74
1.75
1.76
1.77
1.78
1.79
X=1.77
Y=0
The container hits the ground 2.5 seconds after it
is dropped.
7. P 5 0.00256s 2
A pressure of 30 lb/ft2
is produced by a wind
speed between 108.2 and
108.3 miles per hour.
Mixed Review for TEKS (p. 274)
1. B;
416 5 (24 1 2x)(20 1 2x) 2 (24)(20)
0 5 0.00256s 2 2 30
Y1
-.1402
-.0848
-.0295
.02596
.08143
.13696
.19254
Y1
2.1136
1.5584
1
.4384
-.1264
-.6944
-1.266
Area of border 5 Area of frame 2 Area of mirror
30 5 0.00256s 2
X
108
108.1
108.2
108.3
108.4
108.5
108.6
X=108.3
The table feature of the graphing calculator must be set so
that the x-values start at 1 and increase in increments of
0.01. Scroll through the table to ﬁnd the time x at which
the height y of the container is 0 feet.
416 5 480 1 88x 1 4x 2 2 480
0 5 4x 2 1 88x 2 416
0 5 4(x 126)(x 2 4)
x 1 26 5 0
Zero
X=108.25318
Y=0
A pressure of 30 lb/ft2 is
produced by a wind
miles per hour.
or
x2450
x 5 226 or
x54
Reject the negative value, 226. The greatest possible
width of the border is 4 inches.
Algebra 2
Worked-Out Solution Key
201
Chapter 4,
continued
2. H;
h 5 216t 2 1 h0
7.
0 5 216t 1 20
2
672 5 336 1 42x 1 8x 1 x 2
20
0 5 x 2 1 50x 2 336
Î
}
0 5 (x 2 6)(x 1 56)
20
t56 }
16
t 5 61.12
Time must be positive, so the pinecone takes about
1.12 seconds to hit the ground.
3. C;
y 5 20.0035x(x 2 143.9)
y 5 20.0035(x 2 0)(x 2 143.9)
The x-intercepts are 0 and 143.9, so the cannon shoots
143.9 feet.
4. G;
x2650
or
x56
or
x 1 56 5 0
Lesson 4.6
4.6 Guided Practice (pp. 275–279)
1. x 2 5 213
2. x 2 5 238
}
}
x 5 6Ï213
x 5 6Ï238
x 5 6i Ï13
x 5 6i Ï38
}
}
3. x 2 1 11 5 3
4
3
}5}
x 2 5 28
4
w 5 }3h
2
1 2
4
3
h 1 }h
x 2 5 228
}
}
x 5 6Ï28
x 5 6Ï228
x 5 6i Ï 8
x 5 6i Ï 28
x 5 62i Ï 2
x 5 62i Ï 7
}
}
5 225
}
5. 3x 2 2 7 5 231
16
h2 1 }
h 2 5 225
9
6. 5x 2 1 33 5 3
3x 5 224
5x 2 5 230
x 2 5 28
x 2 5 26
2
25
9
} h 2 5 225
h 5 81
4. x 2 2 8 5 236
}
h 2 1 w 2 5 152
2
}
h 5 6Ï81
}
x 5 6Ï28
x 5 6Ï26
x 5 6i Ï 8
x 5 6i Ï6
}
}
}
}
x 5 62i Ï2
h 5 69
Height must be positive, so the height of the screen
is 9 inches.
7. (9 2 i) 1 (26 1 7i) 5 [9 1 (26)] 1 (21 1 7)i
5 3 1 6i
8. (3 1 7i) 2 (8 2 2i) 5 (3 2 8) 1 [7 2 (22)]i
5. A;
y 5 23x 2 18x 2 25
2
b
18
x 5 2}
5 } 5 23 2a
2(23)
5 25 1 9i
9. 24 2 (1 1 i) 2 (5 1 9i) 5 [(24 2 1)(i)] 2 (5 1 9i)
y 5 23(3) 2 18(23) 2 25 5 2 5 (25 2 i) 2 (5 1 9i)
The function y 5 23x 2 2 18x 2 25 has a vertex
of (23, 2).
5 210 2 10i
2
6. Let x 5 \$.25 price increase.
Let R(x) 5 daily revenue.
Daily revenue (dollars) 5
Price (dollars/slice) + Number sold (slices)
5 (25 2 5) 1 (21 2 9)i
10. 5 1 3i 1 (27i) 5 5 2 4i
The impedance of the circuit is 5 2 4i ohms.
11. i(9 2 i) 5 9i 2 i 2
5 9i 2 (21)
R(x) 5 (2 1 0.25x) + (80 2 5x)
5 9i 1 1
R(x) 5 160 2 10x 1 20x 2 1.25x 2
5 1 1 9i
R(x) 5 21.25x 1 10x 1 160
2
10
b
x 5 2}
5 2}
54
2a
2(21.25)
When x 5 4, R is maximized.
The pizza shop should charge \$2 1 4(\$.25), or \$3 a slice
to maximize proﬁt.
202
x 5 256
Reject the negative value, 256. The value of x is 6 feet.
Algebra 2
Worked-Out Solution Key
12. (3 1 i)(5 2 i) 5 15 2 3i 1 5i 2 i 2
5 15 1 2i 2 (21)
5 15 1 2i 1 1
5 16 1 2i
t2 5 }
16
2
New width
(ft)
+
2(42)(8) 5 (42 1 x)(8 1 x)
16t 2 5 20
w
h
New area
New length
5
(sq ft)
(ft)
```
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