Document 1803765

Chapter 4,
continued
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Lesson 4.5
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4.5 Guided Practice (pp. 266–269)
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1. Ï 27 5 Ï 9 + Ï 3 5 3Ï 3
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29 1 Ï 7
R(x) 5 210 + 5(x 2 14)(x 1 10)
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R(x) 5 210(x 2 14)(5x 1 50)
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4
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8 2 Ï3 8 1 Ï3
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4 1 Ï 11 4 2 Ï11
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R(x)
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or
x 5 25 or
9.
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3
Algebra 2
Worked-Out Solution Key
197
Chapter 4,
continued
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16
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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
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22. s 2 5 169
2. To “rationalize the denominator” of a quotient containing
square roots means to eliminate the radical from the
denominator.
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8. 5Ï 2 4 + 3Ï 10 5 15Ï 240 5 151 Ï 16 + Ï 15 2
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4 1 Ï5 4 2 Ï5
16 2 4Ï5 1 4Ï5 2 5
4 1 Ï5
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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
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x 2 1 5 6Ï2
t 2 5 140
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5 2 Ï6
1
5 2 Ï6
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5 1 Ï6 5 2 Ï6
25 2 5Ï6 1 5Ï6 2 6
5 1 Ï6
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5 2 Ï6
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19
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x 5 6Ï100
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t 5 6Ï 140
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5 21 2 Ï 3
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30. } 1 8 5 15
20
2 1 2Ï3
2 1 2Ï 3
2
1 1 Ï3
2
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22
1 2 Ï3 1 1 Ï3
1 1 Ï3 2 Ï3 2 3
1 2 Ï3
4Ï2 2 Ï 10
x2
25
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p 5 6Ï 112
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6. Ï 3 + Ï 27 5 Ï 81 5 9
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s 5 6 13
x 5 6 Ï84
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24. x 2 5 84
5. Ï 150 5 Ï 25 + Ï 6 5 5Ï 6
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a 5 65 Ï2
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23. a 2 5 50
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t 5 6Ï 4 + Ï35
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Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
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x 5 6 Ï 81
expression.
3. Ï 28 5 Ï 4 + Ï 7 5 2Ï 7
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5x2 5 405
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x 2 5 81
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number solutions: x 5 Ï 81 , and x 5 2Ï81 .
1. In the expression Ï 72 , 72 is called the radicand of the
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21. Because 81 > 0, the equation x 2 5 81 has two real-
Skill Practice
198
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20. The expression was not completely simplified.
4.5 Exercises (pp. 269–271)
14.
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230 5 216t 2
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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
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x 5 4 6 2 5 6, 2
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34. C; 3(x 1 2)2 1 4 5 13
40
16
} 5 t2
Ï
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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
}
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x 1 2 5 6}
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}
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x 2 4 5 62
0 5 216t2 1 40
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x1256 }
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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
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
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
36. Sample answer:
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}
} + }
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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
The radius of the circular lot should be about 5.6 feet.
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
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Ï
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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
ground about 1.4
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
your friend catches it.
Y=0
The ball is in the air 1.25
seconds before your
friend catches it.
10. h 5 16t 2 1 h0
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
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
speed of about 108.3
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
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
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)