COLLEGE ALGEBRA I (MATH 006)
FALL 2013
——PRACTICE FOR EXAM II——
SOLUTIONS
1. Solve the following equations.
y
y
2 −2 = 3
Solution:
6( y2 − 2) = 6( y3 )
3y − 12 = 2y
y = 12
2. Solve the following equations.
1
1
1
10
z − 2z − 5z = z+1
Solution:
10
10z(z + 1)( z1 − 2z1 − 5z1 ) = 10z(z + 1)( z+1
)
10(z + 1) − 5(z + 1) − 2(z + 1) = 100z
10z + 10 − 5z − 5 − 2z − 2 = 100z
3z + 3 = 100z
97z = 3
3
z = 97
1
3. Find all real solutions of the following equations.
(a) x4 − 16 = 0
(b) 3(x − 3)3 = 375
5
(c) 2x 3 + 64 = 0
Solution:
5
(a) x4 − 16 = 0 (b) (x − 3)3 = 125 (c) 2x 3 = −64
√
√
5
x = ± 4 16
x − 3 = 3 125
x 3 = −32
3
5
x = ±2
x=3+5
x = (−32)
q
x=8
x = ( 5 (−32))3
x = (−2)3
x = −8
4. A rental company charges $65 a day and 20 cents a mile
for renting a truck. Michael rents a truck for 3 days, and
his bill comes to $275. How many miles did he drive?
Solution:
x = miles driven
Total Cost=(Per Day Cost)(Number of Days)+(Per Mile
Cost)(Number of Miles)
275 = 65(3) + .20x
.20x = 275 − 195 = 80
80
x = .20
= 400 miles
2
5. Phyllis invested $12, 000, a portion earning a simple interest
rate of 4 21 % per year and the rest earning a rate of 4%
per year. After 1 year the total interest earned on these
investments was $525. How much money did she invest at
each rate?
Solution:
x = investment at 4 12 %
12000 − x = investment at 4%
525 = .045x + (12000 − x)(.04)
525 = .045x + 480 − .04x
.005x = 45
x = 9000
$9000 at 4 12 % and $3000 at 4%.
6. Solve the following equations by factoring.
(a) x2 + x = 12
(b) 3x2 − 5x − 2 = 0
Solution:
(a) x2 + x − 12 = 0
(b) 3x2 − 6x + x − 2 = 0
(x + 4)(x − 3) = 0
3x(x − 2) + (x − 2) = 0
x = −4, x = 3
(3x + 1)(x − 2) = 0
3x = −1
x = − 31 , x = 2
3
7. Solve the equation by completing the square.
(a) x2 − 6x − 11 = 0
(b) 2x2 + 8x + 1 = 0
Solution:
(a) x2 − 6x = 11
(b) x2 + 4x + 12 = 0
x2 − 6x + 9 = 11 + 9
x2 + 4x = − 12
(x − 3)2 =√20
x2 + 4x + 4 = − 12 + 82
x−3=√
± 20
(x + 2)2 =q72
x = 3 ± √20
x + 2 = ±q 72
x = −2 ± √ 72
x=3±2 5
x = −2 ± 214
8. Find real solutions of the following equation. (Quadratic
Formula)
3x2 + 6x − 5 = 0
Solution: √
√
−6± 36−4(3)(−5)
−6± 36+60
x =
=
6
√ 6
√
√
−6± 96
4 6
2 6
=
=
−1
±
=
−1
±
6
6
3
9. Use the discriminant to determine the number of real solutions of each equation. Do not solve.
(a) x2 − 6x + 1 = 0
(b) x2 + 2.20x + 1.21 = 0
(c) 4x2 + 5x +
13
8
=0
Solution:
(a) b2 − 4ac = 36 − 4(1)(1) = 36 − 4 = 32 > 0
2 real solutions
(b) b2 − 4ac = (2.2)2 − 4(1)(1.21) = 4.84 − 4.84 = 0
1 real solution
4
(c) b2 − 4ac = 25 − 4(4)( 13
8 ) = 25 − 26 = −1 < 0
no real solution
10. Find all real solutions of the equation.
x3 − 5x2 − 2x + 10 = 0
Solution:
x2 (x − 5) − 2(x − 5) = 0
(x2 − 2)(x − 5) = 0
x2 − 2 √
= 0, or x − 5 = 0
x = ± 2, or x = 5
11. Find all real solutions of the equation.
1
1
5
x−1 + x+2 = 4
Solution:
1
1
4(x − 1)(x + 2)( x−1
+ x+2
) = 4(x − 1)(x + 2) 45
4(x + 2) + 4(x − 1) = 5(x − 1)(x + 2)
4x + 8 + 4x − 4 = 5(x2 + x − 2)
8x + 4 = 5x2 + 5x − 10
5x2 − 3x − 14 = 0
5x2 − 10x + 7x − 14 = 0
5x(x − 2) + 7(x − 2) = 0
(5x + 7)(x − 2) = 0
5x + 7 = 0, or x − 2 = 0
x = − 75 , or x = 2
5
12. √
Find all real solutions of the equation.
2x + 1 + 1 = x
Solution:
√
√2x + 1 =2 x − 1
( 2x + 1) = (x − 1)2
2x + 1 = x2 − 2x + 1
x2 − 4x = 0
x(x − 4) = 0
x = 0,qor x = 4
Since 2(0) + 1 + 1 6= 0, x = 0 is not a solution.
x = 4 is the only solution.
13. Find all real solutions of the equation.
1 2
1
( x+1
) − 2( x+1
)−8=0
Solution:
1
Let r = x+1
r2 − 2r − 8 = 0
(r − 4)(r + 2)
r = 4, or r = −2
1
1
x+1 = 4, or x+1 = −2
x + 1 = 41 , or x + 1 = − 12
x = 14 − 44 , or x = − 12 − 22
x = − 34 , or x = − 23
6
14. Find all real solutions of the equation.
2
4
x 3 − 5x 3 + 6 = 0
Solution:
2
Let r = x 3
r2 − 5r + 6 = 0
(r − 3)(r − 2)
r = 3, or r = 2
2
2
x 3 = 3, or x 3 = 2
3
3
x = ±3√2 , or x = ±2 2√
x = ± √27, or x = ± √8
x = ±3 3, or x = ±2 2
15. Solve the linear inequality.
(a) 2x − 5 > 3
(b) 7 − x ≥ 5
Solution:
(a) 2x > 8 (b) −x ≥ −2
x>4
x≤2
16. Solve the linear inequality.
−2 < 8 − 2x ≤ −1
Solution:
−10 < −2x ≤ −9
5 > x ≥ 92
9
2 ≤x <5
7
17. Solve the nonlinear inequality.
x2 − 3x − 18 ≤ 0
Solution:
(x − 6)(x + 3) ≤ 0
Zeros: x = 6, x = −3
A
B
C
-3
6
A
B
Test Point(t)
−4 0
(t − 6)(t + 3)
10 −18
Inequality Satisfied? No Yes
Solution: −3 ≤ x ≤ 6, or [−3, 6]
C
7
10
No
18. Solve the nonlinear inequality.
2x+1
x−5 ≤ 3
Solution:
3(x−5)
2x+1
x−5 − x−5 ≤ 0
2x+1−3x+15
≤0
x−5
16−x
x−5 ≤ 0
Zeros: numerator x = 16, denominator x = 5
A
B
5
C
16
A
B
C
Test Point(t)
0
10 20
16
4
16−t
6
−5
− 15
t−5
5
Inequality Satisfied? Yes No Yes
Solution: (−∞, 5) ∪ [16, ∞)
8
19. Solve the equation.
3 |x + 5| + 6 = 15
Solution:
3 |x + 5| = 9
|x + 5| = 3
x + 5 = −3 or x + 5 = 3
x = −8
x = −2
20. Solve each inequality.
a) |x − 5| ≤ 3
b) |x + 5| ≥ 3
Solution:
a) −3 ≤ x − 5 ≤ 3 b) x + 5 ≤ −3 or x + 5 ≥ 3
2≤x≤8
x ≤ −8
x ≥ −2
9