Math 140
Quiz 1 - Summer 2004
Solution Review
Math 140
Quiz 1 - Summer 2004
Solution Review
Math 140
Quiz 1 - Summer 2004
Solution Review
(Small white numbers next to problem number
represent its difficulty as per cent getting it wrong.)
Problem 1
(3)
Solve the equation: -7.3q + 1.9 = -59.7 – 1.7q.
-7.3q + 1.7q = -59.7 – 1.9
-5.6q = -61.6
q = 11
Thus, the answer is E).
Problem 2
(69)
Divide and simplify. Assume that all variables
represent positive real numbers.
56 x 5 y 6
2 y4
2y
4
55
28 x y
2y
44
22
 4x y
4
2
7 x  2x y 7 x
2
Problem 3
(38)
Simplify the radicals and combine any like terms.
Assume all variables represent positive real
numbers.
19•21/3 - 3•(2 • 27)1/3 =
19•21/3 - 3•21/3(33)1/3 =
19•21/3 - 9•21/3 =
Problem 4
(79)
Perform the indicated operation and
simplify:
4
+ 2w .
w-2
2-w
Problem 5
Rationalize the denominator:
_______________
__________
(101/2)2 - 32
(55)
10
.
Problem 6
(55)
Use rational exponents to simplify the radical.
Assume that all variables represent positive
numbers.
(9•9)1/12 = (34)1/12 =
34/12 = 31/3 =
Problem 7
Solve the equation:
(52)
.
Problem 7 Continued
Solve the equation:
.
Alternate approach: Multiply by LCD = 12x. Then,
12x – 7(12x)/(3x) = (10/4)(12x)
12x – 28 = 30x
-18x = 28
x = -28/18 = -14/9
Problem 8
(66)
Solve the equation by a u-substitution and factoring.
x4 + x2 – 2 = 0
Let u = x2 . Then the equation is
u2 + u – 2 = 0
There are only two factoring possibilities:
(u - 1) (u + 2) & (u + 1) (u - 2).
But only the combination (u + 2) (u - 1) works.
u2 + u – 2 = (u + 2) (u - 1) = 0 => u = 1 or -2.
Since u= x2 > 0, drop –2 case & deduce x2 = u = 1.
Hence, x2 – 1= (x + 1) (x - 1) = 0 => x = -1 or 1.
Problem 9
(62)
Use radical notation to write the expression.
Simplify if possible:
.
Note: (-1) = (-1)3 & 512 = 29. Thus,
-512 x12 = (-1)3 29 x12 =>
(-512 x12 )1/3 = [(-1)3 29 x12 ]1/3 =
(-1)3/3 29/3 x12/3 = - 23 x4 = - 8x4
Problem 10
(41)
A rectangular carpet has a perimeter of 236
inches. The length of the carpet is 94 inches
more than the width. What are the dimensions of
the carpet?
Let W = width & L = length = W + 94
Perimeter = 2L + 2W = 236
2(W + 94) + 2W = 236
4W=236-188=48 => W=12" & L=106"
Problem 11
(38)
Solve by completing the square: x2 + 8x = 3.
x2 + 8x + (8/2)2 = 3 + (8/2)2
x2 + 8x + 16 = (x + 4)2 = 19
x + 4 = 191/2 or x + 4 = -191/2
x = -4 + 191/2 or x = -4 - 191/2
{-4±
}
Problem 12
(31)
Solve the equation: 18n2 + 78n = 0.
18n2 + 78n = 0
6n(3n + 13) = 0
n = 0 or (3n + 13) = 0
n = 0 or n = -13/3
{-13/3, 0}
Problem 13
(34)
Solve the equation by factoring: x3 + 6x2 - 7x = 0.
x(x2 + 6x - 7) = 0
x(x + 7) (x - 1) = 0
x = 0 or x + 7 = 0 or x - 1 = 0
x = 0 or x = - 7
{-7, 0, 1}
or x = 1
Problem 14
(69)
The manager of a coffee shop has one type of coffee
that sells for $10 per pound (lb) and another type
that sells for $15/lb. The manager wishes to mix 40
lbs of the $15 coffee to get a mixture that will sell
for $14/lb. How many lbs of the $10 coffee should
be used?
Let t = amt of $10/lb &
To have value equal:
f = amt of $15/lb = 40
10t +15f = 14(40+t).
10t +15(40) = 560+ 14t or 10t +600 – 14t = 560
-4t = -40 => t = 10 pounds
Problem 15
(38)
Write each expression in interval notation.
Graph each interval.
x>6
Recall rules: + => Open(+) right/left(-) end
Note: x > 6 => 6 < x so only left end is determined.
Open end
Left side:
a < x => (a,
Closed end
a < x => [a,
(6, ............................)
5
(
6 7 8 9 10
Problem 16
(21)
Write each expression in interval notation.
Graph each interval.
-2 < x < 1
Recall notation rules:
Open end
Closed end
Left side:
a < x => (a,
a < x => [a,
Right side:
x < a => , a)
(-2,
1]
(
]
-3 -2 -1 0 1 2
x < a => , a]
Problem 17
Solve the equation.
(48)
p  5 p  81  p  4
2
p2 - 5p + 81 = (p + 4)2 =
Always check when
p2 + 8p + 16
squaring radical equations
-13p + 65 = 0
since spurious roots can be
introduced. Check:
-13p = -65
2
Does
5

5
(
5
)

81

5

4
?
p=5
_______________________YES!
_ {5} is solution set.
Problem 18
(17)
Solve the equation: |5m + 4| + 8 = 10 .
5m + 4 = 2
or 5m + 4 = -2
5m = -2
or
5m = -6
m = -2/5
or
m = -6/5
{-2/5, -6/5}
Problem 19
Simplify the complex fraction.
49 y 2  16 x 2
7 4


xy
x y
(
)
(45)
1
(7 y  4 x)(7 y  4 x)
xy


xy
(7 y  4 x )
7 y  4x
Problem 20
(41)
Solve the inequality. Write answer in interval notation.
|r + 4| > 2
r+4>2
or
r + 4 < -2
r>2-4
or
r < -2 - 4
r>-2
or
r < -6
(-, -6] or
[-2, )