Version 028 – Homework 1 - PreCalc Review – Helleloid – (58250)
1
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print-out
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have
19
questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
Welcome to Quest.
This homework over-
laps #2 and #3. Use it as a precalculus review
and diagnostic. Good luck with M408K.
001
10.0 points
Rewrite the expression
radicalBig
a
+
√
b .
1.
(
a
+
b
1
/
2
)
1
/
2
correct
2.
a
1
/
2
+
b
1
/
4
3.
a
+
b
4.
a
+
b
1
/
2
5.
a
+
b
1
/
4
Explanation:
Since
√
x
=
x
1
/
2
, we see that
radicalBig
a
+
√
b
= (
a
+
b
1
/
2
)
1
/
2
.
002
10.0 points
Rationalize the numerator of
√
x
+ 4
−
√
x
−
3
x
.
1.
7
x
(
√
x
+ 4 +
√
x
−
3)
correct
2.
1
x
(
√
x
+ 4 +
√
x
−
3)
3.
x
√
x
+ 4 +
√
x
−
3
4.
7
x
(
√
x
+ 4
−
√
x
−
3)
5.
7
x
√
x
+ 4
−
√
x
−
3
Explanation:
By the difference of squares,
(
√
x
+ 4
−
√
x
−
3)(
√
x
+ 4 +
√
x
−
3)
= (
√
x
+ 4)
2
−
(
√
x
−
3)
2
= 7
.
Thus, after multiplying both the numerator
and the denominator in the given expression
by
√
x
+ 4 +
√
x
−
3
,
we obtain
7
x
(
√
x
+ 4 +
√
x
−
3)
.
003
10.0 points
Simplify the expression
parenleftBig
xy
−
3
√
z
parenrightBig
6
÷
parenleftBig
y
2
x
1
/
3
z
−
3
parenrightBig
12
as much as possible, leaving no negative ex-
ponents and no radicals.
1.
x
10
y
42
z
39
correct
2.
x
2
y
6
z
33
3.
x
10
y
6
z
33
4.
x
10
y
42
z
39
5.
x
10
z
39
y
42
Explanation:
By the Laws of Exponents
parenleftBig
xy
−
3
√
z
parenrightBig
6
=
x
6
y
18
z
3
,
while
parenleftBig
y
2
x
1
/
3
z
−
3
parenrightBig
12
=
y
24
z
36
x
4
.
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Version 028 – Homework 1 - PreCalc Review – Helleloid – (58250)
2
Consequently, the given expression can be
rewritten as
x
6
y
18
z
3
×
x
4
y
24
z
36
=
x
10
y
42
z
39
.
004
10.0 points
Simplify the expression
f
(
x
) =
5 +
15
x
−
4
4 + 60
parenleftBig
x
x
2
−
16
parenrightBig
as much as possible.
1.
f
(
x
) =
x
−
4
x
−
16
2.
f
(
x
) =
5
4
parenleftBig
x
+ 4
x
+ 16
parenrightBig
correct
3.
f
(
x
) =
5
4
parenleftBig
x
+ 4
2
x
+ 16
parenrightBig
4.
f
(
x
) =
x
+ 4
x
−
16
5.
f
(
x
) =
5
4
parenleftBig
x
−
4
x
+ 16
parenrightBig
Explanation:
After bringing the numerator to a common
denominator it becomes
5
x
−
20 + 15
x
−
4
=
5
x
−
5
x
−
4
.
Similarly, after bringing the denominator to
a common denominator and factoring it be-
comes
4
x
2
−
64 + 60
x
x
2
−
16
=
4(
x
−
1)(
x
+ 16)
x
2
−
16
.
Consequently,
f
(
x
) =
5 +
15
x
−
4
4 + 60
parenleftBig
x
x
2
−
16
parenrightBig
=
5
x
−
5
4(
x
−
1)(
x
+ 16)
parenleftBig
x
2
−
16
x
−
4
parenrightBig
.
On the other hand,
x
2
−
16 = (
x
+ 4)(
x
−
4)
.
Thus, finally, we see that
f
(
x
) =
5
4
parenleftbigg
x
+ 4
x
+ 16
parenrightbigg
.
005
10.0 points
Find the solution set of the inequality
x
−
2
x
−
4
≤
x
x
−
1
,
expressing your answer in interval notation.
1.
(
−∞
,
−
2 )
∪
(1
,
4)
2.
(
−∞
,
−
2 ]
∪
(1
,
4)
correct
3.
(
−
2
,
1)
∪
(4
,
∞
)
4.
[
−
2
,
1)
∪
(4
,
∞
)
5.
(
−∞
,
−
2 ]
∪
(4
,
∞
)
Explanation:
To solve the inequality
x
−
2
x
−
4
≤
x
x
−
1
we first bring all the terms to one side:
x
−
2
x
−
4
−
x
x
−
1
≤
0
,
and then bring the left hand side to a common

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- Fall '08
- schultz
- Calculus, Quadratic equation, PreCalc Review
-
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