218 Ï

Chapter 4,
continued
5. 3x 2 1 12x 1 12 5 0
Lesson 4.8
b 2 2 4ac 5 122 2 4(3)(12) 5 0
4.8 Guided Practice (pp. 293–295)
One real solution:
x 2 5 6x 2 4
}
}
212 6 Ï 0
2b 6 Ïb2 2 4ac
5}
x 5 }}
2a
2(3)
x 2 2 6x 1 4 5 0
}
2b 6 Ïb 2 2 4ac
x 5 }}
2a
212
5 22
5}
6
}}
2(26) 6 Ï(26)2 2 4(1)(4)
x 5 }}}
2(1)
8x 2 9x 1 11 5 0
2
}
6 6 Ï 20
x5}
2
b2 2 4ac 5 (29)2 2 4(8)(11) 5 2271 < 0
Two imaginary solutions:
}
6 6 2Ï 5
x5}
2
}
}
The solutions
are x 5 3 1 Ï 5 ø 5.24 and
}
x 5 3 2 Ï5 ø 0.76.
4x 2 2 10x 5 2x 2 9
9
7.
7x 2 2 2x 5 5
7x 2 2 2x 2 5 5 0
b2 2 4ac 5 (22)2 2 4(7)(25) 5 144 > 0
}
2b 6 Ïb 2 4ac
x 5 }}
2a
2
Two real solutions:
}
}
2b 6Ïb2 2 4ac
}}
2(22) 6 Ï144
x 5 }}
5 }}
2a
2(7)
2(212) 6 Ï(212)2 2 4(4)(9)
x 5 }}}
2(4)
2 6 12
}
1
5
6
5}
6 }7 5 1, 2}7
5}
7
14
12 6 Ï 0
x5}
8
8. 4x 2 1 3x 1 12 5 3 2 3x
3
x 5 }2
3
The solution is }2.
4x 2 1 6x 1 9 5 0
b2 2 4ac 5 62 2 4(4)(9) 5 2108 < 0
Two imaginary solutions:
7x 2 5x 2 2 4 5 2x 1 3
}
}
2b 6Ï b2 2 4ac
2a
26 6 Ï 2108
2(4)
x 5 }} 5 }}
25x 1 5x 2 7 5 0
2
}
2b 6 Ïb2 2 4ac
x 5 }}
2a
}
26 6 i6Ï 3
}
9.
3x 2 5x 2 1 1 5 6 2 7x
25x 2 1 10x 2 5 5 0
One real solution:
}
}
}
2b 6Ï b2 2 4ac
2a
25 6 i Ï115
x5}
210
210 6 Ï 0
2(25)
x 5 }} 5 }
210
}
51
5}
210
5 6 i Ï115
x5}
10
}
}
3i Ï 3
b 2 2 4ac 5 102 2 4(25)(25) 5 0
25 6 Ï2115
x 5 }}
210
5 1 i Ï115
3
5}
5 2}4 6 }
8
4
}}
25 6 Ï52 2 4(25)(27)
x 5 }}
2(25)
}
5 2 i Ï115
and }
.
The solutions are }
10
10
10. h 5 216t 2 1 v0 t 1 h0
3 5 216t 2 1 50t 1 4
0 5 216t 2 1 50t 1 1
4. 2x 2 1 4x 2 4 5 0
b 2 2 4ac 5 42 2 4(2)(24) 5 48 > 0
Two real solutions:
}}
250 6 Ï502 2 4(216)(1)
t 5 }}
2(216)
}
}
}
24 6 Ï 48
2b 6Ïb 2 2 4ac
5}
x 5 }}
2a
2(2)
}
24 6 4Ï3
5}
4
}
5 21 6 Ï3 ø 0.73, 22.73
218
}
Ï271
5}
6 i}
16
16
4x 2 2 12x 1 9 5 0
3.
}
}
2(29) 6 Ï2271
2b 6 Ïb2 2 4ac
5 }}
x 5 }}
2a
2(8)
x 5 3 6 Ï5
2.
8x 2 5 9x 2 11
6.
Algebra 2
Worked-Out Solution Key
250 6 Ï2564
t 5 }}
232
t ø 20.02 or t ø 3.14
Reject the solution 20.02 because the ball’s time in the
air cannot be negative. So, the ball is in the air for about
3.14 seconds.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1.
Chapter 4,
continued
7. 8w 2 2 8w 1 2 5 0
4.8 Exercises (pp. 296–299)
}}
2(28) 6 Ï(28)2 2 4(8)(2)
w 5 }}}
2(8)
Skill Practice
1. You can use the discriminant of a quadratic equation to
}
8 6 Ï0
determine the equation’s number and type of solutions.
w5}
16
2. Sample answer:
8
w5}
16
When hitting a baseball with a bat, you would need to use
the model that accounts for initial vertical velocity because
the baseball is launched, not dropped.
1
w 5 }2
3. x 2 2 4x 2 5 5 0
1
}}
The solution is }2.
2(24) 6 Ï(24)2 2 4(1)(25)
x 5 }}}
2(1)
8. 5p 2 2 10p 1 24 5 0
}
4 6 Ï 36
x5}
2
}}
2(210) 6 Ï(210)2 2 4(5)(24)
p 5 }}}
2(5)
466
x5}
2
}
10 6 Ï2380
p5}
10
x5263
}
x 5 5, 21
10 6 2i Ï95
p5}
10
The solutions are 5 and 21.
}
Ï95
4. x 2 2 6x 1 7 5 0
p 5 1 6 i}
5
}
}}
Ï95
The solutions are 1 1 i }
and 1 2 i }
.
5
5
}
6 6 Ï8
9. 4x 2 2 8x 1 1 5 0
x5}
2
}}
2(28) 6 Ï(28)2 2 4(4)(1)
2(4)
}
6 6 2Ï 2
x5}
2
x 5 }}}
}
}
8 6 Ï48
x 5 3 6 Ï2
x5}
8
}
The solutions are x 5 3 1 Ï 2 ø 4.41 and
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
}
Ï95
2(26) 6 Ï (26)2 2 4(1)(7)
x 5 }}}
2(1)
}
8 6 4Ï3
}
x5}
8
x 5 3 2 Ï2 ø 1.59.
}
Ï3
5. t 2 1 8t 1 19 5 0
x516}
2
}}
28 6 Ï82 2 4(1)(19)
2(1)
}
Ï3
t 5 }}
ø 1.87 and
The solutions are x 5 1 1 }
2
}
}
Ï3
28 6 Ï212
t5}
x512}
ø 0.13.
2
2
}
28 6 2i Ï3
t5}
2
10. 6u 2 1 4u 1 11 5 0
}}
}
t 5 24 6 i Ï 3
}
}
The solutions are 24 1 i Ï 3 and 24 2 i Ï3 .
24 6 Ï 42 2 4(6)(11)
u 5 }}
2(6)
}
24 6 Ï 2248
u 5 }}
12
6. x 2 2 16x 1 7 5 0
}}
2(216) 6 Ï(216)2 2 4(1)(7)
x 5 }}}
2(1)
}
}
24 6 2i Ï 62
u5}
12
1
16 6 Ï228
x5}
2
}
Ï62
u 5 2}3 6 i }
6
}
1
16 6 2Ï 57
x5}
2
}
Ï62
1
}
Ï62
and 2}3 2 i }
.
The solutions are 2}3 1 i }
6
6
}
x 5 8 6 Ï57
}
The solutions are x 5 8 1 Ï 57 ø 15.55
}
and x 5 8 2 Ï 57 ø 0.45.
Algebra 2
Worked-Out Solution Key
219
Chapter 4,
continued
11. 3r 2 2 8r 2 9 5 0
23y 2 5 6y 2 10
16.
}}
2(28) 6 Ï(28)2 2 4(3)(29)
23y 2 2 6y 1 10 5 0
r 5 }}}
2(3)
8 6 Ï 172
r5}
6
}
6 6 Ï156
y5}
26
}
8 6 2Ï 43
r5}
6
}
6 6 2Ï39
y5}
26
}
4
Ï43
r 5 }3 6 }
3
}
Ï39
y 5 21 6 }
3
}
Ï43
4
ø 3.52 and
The solutions are r 5 }3 1 }
3
}
Ï39
}
Ï43
4
r 5 }3 2 }
ø 20.85
3
ø 1.08 and
The solutions are y 5 21 1 }
3
}
Ï39
y 5 21 2 }
ø 23.08.
3
12. A;
2x 2 2 16x 1 50 5 0
}}
2(216) 6 Ï(216)2 2 4(2)(50)
x 5 }}}
2(2)
17.
3 2 8v 2 5v 2 5 2v
25v 2 2 10v 1 3 5 0
}}
2(210) 6 Ï(210)2 2 4(25)(3)
}
13.
}}
2(26) 6 Ï(26)2 2 4(23)(10)
y 5 }}}
2(23)
}
16 6 Ï 2144
x5}
4
v 5 }}}
2(25)
16 6 12i
x5}
4
v5}
210
x 5 4 6 3i
v5}
210
}
10 6 Ï160
}
10 6 4Ï10
}
3w 2 2 12w 5 212
2Ï 10
v 5 21 6 }
5
3w 2 2 12w 1 12 5 0
}
2Ï10
w 2 2 4w 1 4 5 0
ø 0.26 and
The solutions are v 5 21 1 }
5
}}
2(24) 6 Ï(24)2 2 4(1)(4)
}
2Ï10
w 5 }}}
2(1)
v 5 21 2 }
ø 22.26.
5
}
18.
7x 2 5 1 12x 2 5 23x
12x 2 1 10x 2 5 5 0
w52
14.
x 2 1 6x 5 215
}
210 6 Ï340
x5}
24
x 2 1 6x 1 15 5 0
}
}}
210 6 2Ï85
26 6 Ï62 2 4(1)(15)
x 5 }}
2(1)
x5}
24
}
6}
x 5 2}
12
12
2
}
5
26 6 2i Ï6
x5}
2
5
}
}
}
s 2 5 214 2 3s
4x 2 1 3 5 x 2 2 7x
19.
3x 1 7x 1 3 5 0
2
x 5 }}
2(3)
23 6 Ï3 2 4(1)(14)
2
s 5 }}
2(1)
}
27 6 Ï 13
}
23 6Ï247
s5}
2
x5}
6
}
7
23 6 i Ï47
s5}
2
}
7
Ï47
2
}
Ï13
ø 20.57 and
The solutions are x 5 2}6 1 }
6
}
Ï47
2
3
2
}
Ï47
2
The solutions are 2} 1 i } and 2} 2 i }.
Algebra 2
Worked-Out Solution Key
}
Ï13
x 5 2}6 6 }
6
s 5 } 6 i}
220
}}
27 6 Ï 7 2 2 4(3)(3)
}}
3
2
}
Ï85
x 5 2}
2}
ø 21.18.
12
12
The solutions are 23 1 i Ï6 and 23 2 i Ï6 .
23
2
}
Ï85
1}
ø 0.35 and
The solutions are x 5 2}
12
12
x 5 23 6 i Ï 6
s 2 1 3s 1 14 5 0
}
Ï85
5
26 6 Ï224
x5}
15.
}}
210 6 Ï102 2 4(12)(25)
x 5 }}
2(12)
The solution is 2.
7
}
Ï13
ø 21.77.
x 5 2}6 2 }
6
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
4 6 Ï0
w5}
2
Chapter 4,
continued
20. 6 2 2t 2 5 9t 1 15
23. x 2 2 5x 1 10 5 4
22t 2 9t 2 9 5 0
x 2 2 5x 1 6 5 0
2
}}
}}
2(29) 6 Ï(29)2 2 4(22)(29)
2(25) 6 Ï(25)2 2 4(1)(6)
t 5 }}}
2(22)
x 5 }}}
2(1)
}
}
9 6 Ï9
5 6 Ï1
t5}
24
x5}
2
963
561
t5}
24
9
x5}
2
3
t 5 2}4 6 }4
x 5 3, 2
The solutions are 3 and 2.
3
t 5 2}2 , 23
Check: x 2 2 5x 1 6 5 0
(x 2 3)(x 2 2) 5 0
3
The solutions are 2}2 and 23.
x 2 3 5 0 or x 2 2 5 0
x 5 3 or x 5 2
4 1 9n 2 3n 5 2 2 n
2
21.
24. m 2 1 5m 2 99 5 3m
23n 1 10n 1 2 5 0
2
m 2 1 2m 2 99 5 0
}}
210 6 Ï102 2 4(23)(2)
n 5 }}
2(23)
}}
22 6 Ï22 2 4(1)(99)
m 5 }}
2(1)
}
210 6 Ï124
n5}
26
}
22 6 Ï 400
m 5}
2
}
210 6 2Ï31
n5}
26
22 6 20
m5}
2
}
Ï31
5
n 5 }3 6 }
3}
m 5 9, 211
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Ï31
5
The solutions are n 5 }3 1 }
ø 3.52 and
3
}
Ï31
5
n 5 }3 2 }
ø 20.19.
3
The solutions are 9 and 211.
Check: m 2 1 2m 2 99 5 0
(m 2 9)(m 1 11) 5 0
m 2 9 5 0 or m 1 11 5 0
22. z 2 1 15z 1 24 5 232
m 5 9 or m 5 211
z 1 15z 1 56 5 0
2
25.
}}
215 6 Ï152 2 4(1)(56)
s 2s235s
2
s 2 2 2s 2 3 5 0
z 5 }}
2(1)
}}
2(22) 6 Ï(22)2 2 4(1)(23)
2(1)
}
215 6 Ï1
z5}
2
s 5 }}}
}
2 6 Ï16
215 6 1
z5}
2
s5}
2
z 5 27, 28
s5}
2
264
The solutions are 27 and 28.
s 5 3, 21
Check: z 2 1 15z 1 56 5 0
The solutions are 3 and 21.
(z 1 8)(z 1 7) 5 0
z1850
z 5 28
or
or
z1750
z 5 27
Check: s 2 2 2s 2 3 5 0
(s 2 3)(s 1 1) 5 0
s 2 3 5 0 or s 1 1 5 0
s 5 3 or s 5 21
Algebra 2
Worked-Out Solution Key
221
Chapter 4,
continued
26. r 2 2 4r 1 8 5 5r
29. 5p 2 1 40p 1 100 5 25
r 2 9r 1 8 5 0
5p 2 1 40p 1 75 5 0
2
}}
p 2 1 8p 1 15 5 0
2(29) 6 Ï(29)2 2 4(1)(8)
r 5 }}}
2(1)
}}
28 6 Ï82 2 4(1)(15)
p 5 }}
2(1)
}
9 6 Ï 49
r5}
2
}
28 6 Ï 4
967
r5}
2
p 5 23, 25
r 5 8, 1
The solutions are 23 and 25.
The solutions are 8 and 1.
Check: p 2 1 8p 1 15 5 0
Check: r 2 2 9r 1 8 5 0
( p 1 3)( p 1 5) 5 0
(r 2 8)(r 2 1) 5 0
p1350
r 2 8 5 0 or r 2 1 5 0
or p 1 5 5 0
p 5 23 or
r 5 8 or r 5 1
p 5 25
30. 9n 2 2 42n 2 162 5 21n
27. 3x 2 1 7x 2 24 5 13x
9n 2 2 63n 2 162 5 0
3x 2 2 6x 2 24 5 0
n 2 2 7n 2 18 5 0
x 2 2 2x 2 8 5 0
}}
2(27) 6 Ï(27)2 2 4(1)(218)
n 5 }}}
2(1)
}}
2(22) 6 Ï(22)2 2 4(1)(28)
x 5 }}}
2(1)
}
7 6 Ï121
}
7 6 11
5}
n5}
2
2
2 6 Ï 36
x5}
2
n 5 9, 22
266
x5}
2
The solutions are 9 and 22.
x 5 4, 22
Check: n 2 2 7n 2 18 5 0
(n 2 9)(n 1 2) 5 0
The solutions are 4 and 22.
n 5 9 or
(x 2 4)(x 1 2) 5 0
n 5 22
31. x 2 8x 1 16 5 0
2
x 2 4 5 0 or x 1 2 5 0
b 2 2 4ac 5 (28)2 2 4(1)(16) 5 0
x 5 4 or x 5 22
28. 45x 2 1 57x 1 1 5 5
One real solution:
}
}
45x 2 1 57x 2 4 5 0
2(28) 6 Ï0
2b 6 Ïb 2 2 4ac
x 5 }}
5}
54
2a
2(1)
}}
257 6 Ï572 2 4(45)(24)
x 5 }}
2(45)
32. s 2 1 7s 1 11 5 0
}
b 2 2 4ac 5 72 2 4(1)(11) 5 5 > 0
257 6 Ï3969
x 5 }}
90
Two real solutions:
257 6 63
x5}
90
27 6 Ï5
2b 6 Ïb 2 2 4ac
s 5 }} 5 } ø 22.38, 24.62
1
}
2a
4
, 2}3
x5}
15
2(1)
33. 8p 1 8p 1 3 5 0
2
4
1
The solutions are }
and 2}3 .
15
b 2 2 4ac 5 82 2 4(8)(3) 5 232 < 0
Two imaginary solutions:
Check: 45x 2 1 57x 2 4 5 0
}
15x 2 1 5 0 or 3x 1 4 5 0
1
or
x5}
15
}
}
28 6 Ï 232
28 6 4i Ï 2
2b 6 Ïb 2 2 4ac
5}
5}
p 5 }}
2a
16
2(8)
(15x 2 1)(3x 1 4) 5 0
Algebra 2
Worked-Out Solution Key
}
4
x 5 2}3
1
}
Ï2
5 2}2 6 i }
4
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
n 2 9 5 0 or n1 2 5 0
Check: x 2 2x 2 8 5 0
2
222
28 6 2
5}
p5}
2
2
Chapter 4,
continued
34. 24w 2 1 w 2 14 5 0
7r 2 2 5 5 2r 1 9r 2
39.
b 2 4ac 5 1 2 4(24)(214) 5 2223 < 0
22r 2 2r 2 5 5 0
Two imaginary solutions:
b 2 2 4ac 5 (22)2 2 4(22)(25) 5 236 < 0
2
2
}
2
}
Two imaginary solutions:
21 6 Ï2223
2b 6 Ïb2 2 4ac
w 5 }}
5 }}
2a
2(24)
}
2b 6 Ï b 2 2 4ac
r 5 }}
2a
}
21 6 i Ï223
5}
28
1
}
2(22) 6 Ï236
5 }}
2(22)
}
Ï223
5 }8 6 i }
8
2 6 6i
5}
24
35. 5x 1 20x 1 21 5 0
2
b 2 2 4ac 5 20 2 2 4(5)(21) 5 220 < 0
1
Two imaginary solutions:
}
}
220 6 Ï220
2b 6 Ïb 2 2 4ac
x 5 }}
5 }}
2a
2(5)
16t 2 2 7t 5 17t 2 9
40.
16t 2 24t 1 9 5 0
2
}
(4t 2 3)(4t 2 3) 5 0
220 6 2i Ï 5
5}
10
4t 2 3 5 0
}
Ï5
5 22 6 i }
5
3
t 5 }4
8z 2 10 5 z 2 2 7z 1 3
36.
3
5 2}2 6 }2 i
3
The solution is }4.
2z 2 1 15z 2 13 5 0
b 2 2 4ac 5 15 2 2 4(21)(213) 5 173 > 0
7x 2 3x 2 5 85 1 2x 2 1 2x
41.
25x 1 5x 2 85 5 0
2
Two real solutions:
}
x 2 2 x 1 17 5 0
}
215 6 Ï173
2b 6 Ïb 2 2 4ac
z 5 }}
5}
2a
2(21)
}}
2(21) 6 Ï (21)2 2 4(1)(17)
x 5 }}}
2(1)
}
215 6 Ï173
5}
2(21)
}
1 6 Ï267
x5}
2
}
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
15 6 Ï 173
5}
2
}
1 6 i Ï 67
x5}
2
ø 14.08, 0.92
37.
8n 2 2 4n 1 2 5 5n 2 11
1
8n 2 2 9n 1 13 5 0
1
Two imaginary solutions:
}
}
2(29) 6 Ï2335
2b 6 Ïb 2 4ac
n 5 }}
5 }}
2a
2(8)
2
}
9 6 i Ï335
5}
16
1
}
i Ï67
4(x 2 1)2 5 6x 1 2
42.
4(x 2 2x 1 1) 5 6x 1 2
2
4x 2 2 8x 1 4 5 6x 1 2
4x 2 2 14x 1 2 5 0
2x 2 2 7x 1 1 5 0
}
Ï335
6 i}
5}
16
16
}}
2(27) 6 Ï(27)2 2 4(2)(1)
2(2)
x 5 }}}
38. 5x 2 1 16x 5 11x 2 3x 2
}
7 6 Ï 41
8x 2 1 5x 5 0
b 2 2 4ac 5 52 2 4(8)(0) 5 25 > 0
Two real solutions:
}
}
25 6 Ï 25
2b 6 Ïb 2 2 4ac
x 5 }}
5}
2a
2(8)
25 6 5
}
i Ï 67
and }2 2 }
.
The solutions are }2 1 }
2
2
b 2 2 4ac 5 (29)2 2 4(8)(13) 5 2335 < 0
9
}
i Ï67
x 5 }2 6 }
2
x5}
4
}
7 1 Ï 41
ø 3.35 and
The solutions are x 5 }
4
}
7 2 Ï41
ø 0.15.
x5}
4
5
5 0, 2}8
5}
16
Algebra 2
Worked-Out Solution Key
223
continued
25 2 16v 2 5 12v(v 1 5)
43.
46. 1.1(3.4x 2 2.3)2 5 15.5
25 2 16v 5 12v 1 60v
2
11(3.4x 2 2.3)2 5 155
2
228v 2 2 60v 1 25 5 0
23
17
x2}
1}
10 2
5
2
28v 1 60v 2 25 5 0
2
Î155
11
23
17
155
}x 5 } 6Î }
10
5
11
5 23
155
} 6Î } 2
x5}
17 1 10
11
23
5 155
Î}
x5}
6}
34
17 11
}x 2 } 5 6 }
14v 2 5 5 0
or 2v 1 5 5 0
5
v5}
14
}
5
v 5 2}2
or
}
5
5
and 2}2 .
The solutions are }
14
44.
}
3
4
3
2
} y 2 2 6y 5 } y 2 9
6y 2 24y 5 3y 2 36
6y 2 2 27y 1 36 5 0
}}
23
2(29) 6 Ï(29)2 2 4(2)(12)
y 5 }}}
2(2)
19.25 5 8.5(2r 2 1.75)2
}
Î
9 6 i Ï 15
}
}
}
i Ï 15
3
3x 2 1 }2 x 2 4 5 5x 1 }4
7
8
7
}
iÏ2618
7
}}
2(22) 6 Ï(22)2 2 4(12)(219)
x 5 }}}
2(12)
}
2 6 Ï 916
x5}
24
}
2 6 2Ï 229
x5}
24
1 6 Ï 229
x5}
12 }
1 1 Ï229
The solutions are x 5 }
ø 1.34 and
12
48.
4.5 5 1.5(3.25 2 s)2
3 5 (3.25 2 s)2
}
6Ï 3 5 3.25 2 s
}
s 5 3.25 6 Ï3
}
The solutions are s 5 3.25 1 Ï 3 ø 4.98 and
}
s 5 3.25 2 Ï3 ø 1.52.
49. The solutions are imaginary, not real.
3x2 1 6x 1 15 5 0
}}
26 6 Ï62 2 4(3)(15)
x 5 }}
2(3)
}
26 6 Ï 2144
5 }}
6
26 6 12i
5}
6
5 21 6 2i
The solutions are 21 1 2i and 21 2 2i.
224
Algebra 2
Worked-Out Solution Key
}
iÏ 2618
The solutions are }8 1 }
and }8 2 }
.
68
68
12x 2 2 2x 2 19 5 0
ø 21.18.
x5}
12
}
iÏ2618
68
}6}5r
12x 2 1 18x 2 16 5 20x 1 3
}
}
7
4
9
}
7
} 6 i } 5 2r
and }4 2 }
.
The solutions are }4 1 }
4
4
9
77
Î7734
7
1 77
} 6 }i Î } 5 r
8
2 34
}
i Ï 15
}
7 2
6 2}
5 2r 2 }4
34
y 5 }4 6 }
4
i Ï15
2}
5 1 2r 2 }4 2
34
77
y5}
4
9
}
5Ï1705
2}
ø 20.43.
x5}
34
187
47.
}
9 6 Ï 215
y5}
4
1 2 Ï 229
}
5Ï 1705
23
1}
ø 1.78 and
The solutions are x 5 }
34
187
2y 2 2 9y 1 12 5 0
45.
}
5Ï1705
23
6}
x5}
34
187
2
9
}
23
10
17
5
(14v 2 5)(2v 1 5) 5 0
155
5}
11
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 4,
Chapter 4,
continued
50. The quadratic equation must be written in standard form
before applying the quadratic formula.
x 1 6x 1 8 5 2
2
x 1 6x 1 6 5 0
2
}}
26 6 Ï62 2 4(1)(6)
x 5 }}
2(1)
}
26 6 Ï12
x5}
2
}
26 6 2Ï3
x5}
2
}
x 5 23 6 Ï 3
}
The solutions are x 5 23 1 Ï 3 ø 21.27 and
}
x 5 23 2 Ï3 ø 24.73.
51. ax 2 1 bx 1 c 5 0
}
2b 6 Ïb 2 2 4ac
x 5 }}
2a
Mean of solutions:
}
}
2b 1 Ïb 2 2 4ac 1 1 2b 2 Ïb 2 2 4ac 2
}}}
2a
}
2
22b
}
22b
b
2a
5 } 5 } 5 2} , which is the formula for the
2
4a
2a
axis of symmetry.
The axis of symmetry of the graph of y 5 ax 2 1 bx 1 c
is the mean of the x-values of two points that lie on the
graph that have the same y-value.
52. Because there are two x-intercepts, the discriminant
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
is positive.
53. Because there are no x-intercepts, the discriminant
57. x 2 1 8x 1 c 5 0
a. 82 2 4(1)(c) > 0
64 2 4c > 0
24c > 264
c < 16
b. 82 2 4(1)(c) 5 0
64 2 4c 5 0
24c 5 264
c 5 16
c. 82 2 4(1)(c) < 0
64 2 4c < 0
24c < 264
c > 16
58. 2x 2 1 16x 1 c 5 0
a. 162 2 4(21)(c) > 0
256 1 4c > 0
4c > 2256
c > 264
b. 162 2 4(21)(c) 5 0
256 1 4c 5 0
4c 5 2256
c 5 264
c. 162 2 4(21)(c) < 0
256 1 4c < 0
4c < 2256
c < 264
59. 3x 2 1 24x 1 c 5 0
a. 242 2 4(3)(c) > 0
576 2 12c > 0
is negative.
212c > 2576
54. Because there is one x-intercept, the discriminant is zero.
c < 48
55. C;
b. 24 2 4(3)(c) 5 0
2
2x 2 1 5x 1 c 5 0
576 2 12c 5 0
b2 2 4ac 5 52 2 4(2)(c) 5 223
212c 5 2576
25 2 8c 5 223
28c 5 248
c56
c 5 48
c. 24 2 4(3)(c) < 0
2
576 2 12c < 0
56. x 2 2 4x 1 c 5 0
212c < 2576
a. (24)2 2 4(1)(c) > 0
16 2 4c > 0
24c > 216
c<4
b. (24)2 2 4(1)(c) 5 0
16 2 4c 5 0
24c 5 216
c54
c. (24)2 2 4(1)(c) < 0
16 2 4c < 0
24c < 216
c>4
c>4
60. 24x 2 10x 1 c 5 0
2
a. (210)2 2 4(24)(c) > 0
100 1 16c > 0
16c > 2100
25
c > 2}
4
b. (210)2 2 4(24)(c) 5 0
100 1 16c 5 0
16c 5 2100
25
c 5 2}
4
Algebra 2
Worked-Out Solution Key
225
Chapter 4,
continued
c. (210)2 2 4(24)(c) < 0
ax 2 1 bx 1 4 5 0
64.
a(21) 1 b(21) 1 4 5 0
100 1 16c < 0
2
a2b1450
16c < 2100
25
c < 2}
4
61. x 2 2 x 1 c 5 0
1 2
a5b24
1 2
4 2
4
a 2}3 1 b 2}3 1 4 5 0
a. (21) 2 2 4(1)(c) > 0
16
9
4
3
}a 2 }b 1 4 5 0
1 2 4c > 0
4
3
16
9
24c > 21
}(b 2 4) 2 }b 1 4 5 0
1
c < }4
16(b 2 4) 2 12b 1 36 5 0
16b 2 64 2 12b 1 36 5 0
b. (21) 2 4(1)(c) 5 0
2
4b 2 28 5 0
1 2 4c 5 0
4b 5 28
24c 5 21
b57
1
c 5 }4
a5b24572453
c. (21)2 2 4(1)(c) < 0
ax 2 1 bx 1 4 5 0
1 2 4c < 0
3x 2 1 7x 1 4 5 0
24c < 21
ax 2 1 bx 1 4 5 0
65.
a(21 2 i) 1 b(21 2 i) 1 4 5 0
2
1
c > }4
2ia 1 (21 2 i)b 1 4 5 0
62. b2 2 4ac 5 210
2ia 5 24 2 (21 2 i)b
1 2
7
Sample answer: 5 2 4 }4 (5) 5 25 2 35
2
24 2 (21 2 i)b
a 5 }}
2i
5 210
a(21 1 i)2 1 b(21 1 i) 1 4 5 0
7
}x 2 1 5x 1 5 5 0
4
F
ax 2 1 bx 1 4 5 0
63.
24 2 (21 2 i)b
G
1 (21 1 i)b 1 4 5 0
22i }}
2i
a(24)2 1 b(24) 1 4 5 0
4 1 (21 2 i)b 1 (21 1 i)b 1 4 5 0
16a 2 4b 1 4 5 0
(21 2 i 2 1 1 i)b 5 28
16a 5 4b 2 4
22b 5 28
4b 2 4
a5}
16
b54
24 2 (21 2 i)b
24 2 (21 2 i)(4)
5 }}
a 5 }}
2i
2i
24 1 4 1 4i
5}
52
2i
b21
a5}
4
a(3)2 1 b(3) 1 4 5 0
9a 1 3b 1 4 5 0
ax 2 1 bx 1 4 5 0
b21
1 3b 1 4 5 0
9}
4
1
2
2x 2 1 4x 1 4 5 0
66.
9(b 2 1) 1 12b 1 16 5 0
9b 2 9 1 12b 1 16 5 0
21b 5 27
7
b21
24
3
}
}
ax 1 bx 1 4 5 0
1
1
2}3x 2 2 }3x 1 4 5 0
226
Algebra 2
Worked-Out Solution Key
29a 1 (3i)b 1 c 5 0
24a 1 (22i)b 1 c 5 0
ib 5 a
29(ib) 1 (3i)b 1 c 5 0
1
5}
5 4 5 2}3
a5}
4
4
2
a(22i)2 1 b(22i) 1 c 5 0
(5i)b 5 5a
1
b 5 2}3
1
a(3i)2 1 b(3i) 1 c 5 0,
29a 1 (3i)b 1 c 5 24a 1 (22i)b 1 c
b 5 2}
21
2}3 2 1
ax 2 1 bx 1 c 5 0
(29i)b 1 (3i)b 1 c 5 0
(26i)b 1 c 5 0
c 5 6ib
You can see from the equations a 5 ib and c 5 6ib, that
a, b, and c cannot be real numbers.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
22ia 1 (21 1 i)b 1 4 5 0
Chapter 4,
continued
ax 2 1 bx 1 c 5 0
67.
c. Because the x-coordinate of the vertex is 80, the
horizontal distance h the motorcycle has traveled
when it reaches its maximum height is 80 feet.
b
c
x 2 1 }a x 1 }a 5 0
b
1
1
d. y 5 2} (80)2 1 } (80) 1 20 5 30
640
4
c
x2 1 }
a x 5 2}
a
Because the y-coordinate of the vertex is 30, the
motorcycle’s maximum height k above the ground is
30 feet.
x2 1 }
5 2}a 1 1 }
a x 1 1}
2a 2
2a 2
b 2
b
b 2
c
2c
b2
4a
x2 1 1 }
5}
a 1 }2
2a 2
b 2
24ac 1 b 2
4a
b
1x 1 }
2a 2
5}
2
b
1x 1 }
2a 2
5}
2
2
2
71.
S 5 20.000013E 2 1 0.042E 2 21
10 5 20.000013E 2 1 0.042E 2 21
0 5 20.000013E 2 1 0.042E 2 31
}}}
b2 2 4ac
4a
Î
20.042 6 Ï(0.042)2 2 4(20.000013)(231)
E 5 }}}}
2(20.000013)
}
b
}
20.042 6 Ï 0.000152
b 2 2 4ac
4a
x1}
56 }
2
2a
E 5 }}
20.000026
E ø 1141.2 or E ø 2089.57
}
b
Ïb 2 2 4ac
x 5 2}
6}
2a
2a
You would expect to find 10 species of ants at elevations
of 1141.2 meters and 2089.57 meters.
}
2b 6 Ïb2 2 4ac
x 5 }}
2a
72. a. 4* 1 3w 5 900
3w 5 900 2 4*
Problem Solving
4
w 5 300 2 }3*
68. h 5 216t 2 1 v0 t 1 h0
0 5 216t 2 2 50t 1 7
b. *w 5 12,000
}}
2(250) 6 Ï(250) 2 4(216)(7)
2
t 5 }}}
2(216)
12,000
*5}
w
w 5 300 2 }3 1 }
w 2
4 12,000
}
50 6 Ï2948
t5}
232
16,000
w 5 300 2 }
w
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
t ø 23.26 or t ø 0.13
w 2 5 300w 2 16,000
Reject the solution 23.26 because the ball’s time in
the air cannot be negative. So the defensive player’s
teammates have about 0.13 second to intercept the ball
before it hits the ground.
w 2 2 300w 1 16,000 5 0
}}
2(2300) 6 Ï(2300)2 2 4(1)(16,000)
w 5 }}}
2(1)
69. C;
}
300 6 Ï26,000
w 5 }}
2
s 5 858t 1 1412t 1 4982
2
}
50,000 5 858t 2 1 1412t 1 4982
300 6 20Ï65
0 5 858t 1 1412t 2 45,018
2
w ø 230.62 or w ø 69.38
}}}
21412 6 Ï(1412)2 2 4(858)(245,018)
2(858)
t 5 }}}
12,000
230.62
When w ø 230.62: * ø } ø 52.03
}}
21412 6 Ï156,495,520
t 5 }}
1716
12,000
69.38
When w ø 69.38: * ø } ø 172.96
t ø 6.47 or t ø 28.11
Reject the solution 28.11. The number of subscribers
reached 50 million 6 years after 1990, or 1996.
70. a. The motorcycle’s height r when it lands on the ramp is
The possible dimensions of each section are 230.62
feet by 52.03 feet or 69.38 feet by 172.96 feet.
73. x 5 20t
y 5 216t 2 1 21t 1 6
20 feet.
1
1
b. y 5 2}x 2 1 }x 1 20
640
4
}
5 150 6 10Ï65
w 5 }}
2
a.
t
0
0.25
0.5
0.75
1
2}4
b
} 5 80
x 5 2}
5
2a
1
21 2}
640 2
x
0
5
10
15
20
y
6
10.25
12.5
12.75
11
Because the x-coordinate of the vertex is 80, the
distance d between the ramps is 2(80), or 160 feet.
(x, y)
1
(0, 6) (5, 10.25)
(10, 12.5) (15, 12.75)
(20, 11)
Algebra 2
Worked-Out Solution Key
227
Chapter 4,
b.
continued
Lesson 4.9
y
4.9 Guided Practice (pp. 301–303)
16
1. y > x 2 1 2x 2 8
12
Test (0, 0):
y > x 2 1 2x 2 8
0
> 02 1 2(0) 2 8
y
2
8
(0, 0)
21
x
0 > 28 4
0
0
5
10
15
x
20
c. No, the player does not make the free throw. The shot
is too high. It goes over the backboard.
2. ya2x 2 2 3x 1 1
74. a. h 5 216t 1 v0 t 1 921
2
The maximum height of 1081 feet occurs at
the vertex.
v0
v0
b
t 5 2}
5 2}
5}
2a
32
2(216)
v0 2
v0
1 2
(1, 2)
2÷0
21
1 2
v02
v0
ya2x 2 2 3x 1 1
2(1)2 2 3(1) 1 1
2a
1
1 v0 }
1 921 5 1081
h 5 216 }
32
32
2
Test (1, 2):
y
x
3. y < 2x 2 1 4x 1 2
1}
1 921 5 1081
2}
64
32
Test (0, 0):
y < 2x 2 1 4x 1 2
0
< 2(0)2 1 4(0) 1 2
y
v02
} 5 160
0<2
64
v02 5 10,240
2
}
v0 5 6Ï 10,240 5 6101
21
(0, 0)
x
The initial velocity is about 101 feet per second.
}
b. When v0 5 Ï 10,240 :
4. yqx 2
y < 2x 2 1 5
1081 5 216t 2 1 Ï 10,240 t 1 921
}
0 5 216t 2 1 Ï10,240 t 2 160
}
}}}
}
}
}
y
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
}
Ï 1 Ï10,240 22 2 4(216)(2160)
t 5 }}}}
2Ï 10,240 6
2(216)
1
}
}
2Ï 10,240 6 Ï 0
232Ï10
t 5 }}
5}
5 Ï 10 ø 3.16
232
232
The time given by the model is longer than the
time given in the brochure. The model is not
extremely accurate.
21
2x 2 1 2xa
5.
2x 2 1 2x 2 3a0
2x 2 1 2x 2 3 5 0
}}
22 6 Ï22 2 4(2)(23)
x 5 }}
2(2)
Mixed Review for TAKS
75. D;
}
21 6 Ï 7
908 1 908 1 (2x 1 50)8 1 2x8 1 (3x 2 5)8 5 5408
5}
2
7x 1 225 5 540
76. F;
x
}
x
s 1 a 5 725
4s 1 6a 5 3650
2x 2 1 2x 2 3
23
9
21 2 Ï 7
22 }
2
1
21
23
0
}
x
2x 2 1 2x 2 3
0
21 1 Ï7
}
1
23
0
1
2
2
9
}
}
21 2 Ï7
21 1 7
The solution of the inequality is }
axa}
.
2
Ï
2
228
Algebra 2
Worked-Out Solution Key