Intermediate Algebra Diagnostic Test

Intermediate Algebra Diagnostic Test
Mathematics Department
January 26, 2004
Home Page
Instructions: Click the “Begin Quiz” button before you begin your selections. For each
question, click the checkbox containing the “best” answer to the question. When you complete
the quiz, click the “End Quiz” button with your mouse to obtain your results.
You can also obtain corrections to the quiz by clicking the “Correct” button. Answers are
marked according to the following legend.
Legend: A ✔ indicates that the quiz-taker gave the correct response. A ✘ indicates an incorrect
response. In this case, the correct answer is marked with a ●. You can examine solutions by
clicking the correct answer marker ●.
Title Page
◭◭ ◮◮
◭
◮
Page 1 of 100
1. if x = 2 and y = −3, then (xy2 ) 3
x 2 y 5 =
4 6 −6
−12 8
Go Back
Full Screen
Close
Quit

2. x + 1
x − 3 · x 2 − x − 6
x 2 + 5x + 4 =
x + 1
x + 4
x + 2
x − 3
x + 2
x + 4
x + 6
x + 4
x + 3
x − 3
3. If x − 4 x = 5, then x =
5
15 20 25
30 35
4. (2.3 × 10 6 ) × (5.0 × 10 −1 ) =
1.15 × 10 −6 11.5 × 10 −6 2.8 × 10 5
7.3 × 10 5 1.15 × 10 6
5. The inequality 1 − 3x < 7 is equivalent to:
x < −2 x > 6 x < −5
Home Page
Title Page
◭◭ ◮◮
◭
◮
Page 2 of 100
x > −1
6. a2 b
35 · 7
b 3 · 1
a =
x > −2
ab
a
5
5b
a
5a 2
5b 2 b
a 2
5b 3
Go Back
Full Screen
Close
Quit

7. x−4
x −8 =
x 4 x 32 x −4
x −12 x 12
8. If x = −2, then x|x| equals
4 −4 0
−8 8
9. In the figure, the slope of line segment AB is 1/2. Calculate the y-coordinate of point A.
Do not assume that picture is drawn to scale.
y
B(2, 4)
Home Page
Title Page
◭◭ ◮◮
◭
◮
x
Page 3 of 100
A(−3, y)
Go Back
Full Screen
3/4 2/3 3/2
−4/5 5/3
Close
Quit

10. 2a − [2(3a − b) − 4b] =
6a − b 4a − 2b 2a − 6b
−4a + 6b
2a − 4b
11. If 1 x + 1 y = 1 z , then z =
x + y
xy(x + y)
12. (−2ab 2 ) 3 (3ab) =
13. If
x + y
xy
xy
xy
x + y
−5a 3 b 6 −24a 3 b 6 −18ab 2
−18a 4 b 7 −24a 4 b 7
then x =
x + y = a
x − 2y = b,
(a + b)/2 (2a + b)/3 (a + 2b)/3
(a + b)/3 (a − 2b)/3
14. If f(x) = 1 − 2x − x 2 , then f(−2) =
15. √ 3 √ 21 =
9 8 −6
1 3
7 √ 3 3 √ 7
9 √ √
3
7
√
24
Home Page
Title Page
◭◭ ◮◮
◭
◮
Page 4 of 100
Go Back
Full Screen
Close
Quit

16. Which of the following is most likely the graph of 2x − 5y = 10
y
y
x
x
Home Page
Title Page
y
y
◭◭
◮◮
x
x
◭
◮
Page 5 of 100
Go Back
None of the above
17. a b + c d =
Full Screen
a + c
b + d
ac + bd
bd
ac
bd
ad + bc
bd
a + c
bd
Close
Quit

18. x2n+3
x n−2 =
x n+5 x n+1 x (2n+3)/(n−2)
x 3n+1
x 6n
19. Which of the following is most likely the graph of x = y 2
y
y
Home Page
x
x
Title Page
◭◭
◮◮
y
y
◭
◮
Page 6 of 100
x
x
Go Back
Full Screen
None of the above
Close
Quit

20. What is the slope of the line defined by the equation 2x − 5y = 10
5 −5/2 2/5
−1/5 −2
21. log a 2 = 3 is equivalent to
22.
2 a = 3 a = 8 a 2 = 3
a 3 = 2 3 a = 2
a
b − a + b
a − b =
1 −1
b
a − b
a + b
a − b
23. 8x
4 x = 2 2x 4 x
a
a − b
Home Page
Title Page
◭◭ ◮◮
◭
◮
2 x 4
24. One of the roots of x 2 − 7x = 8 is
−9 25 11
−2 8
25. If f(x) = x 2 , then
f(x) − f(2)
x − 2
x + 2 x − 2 2x
2x/(x − 2) x 2
=
Page 7 of 100
Go Back
Full Screen
Close
Quit

26. Which of the following is most likely the graph of y = mx, where m is a negative real
number
y
y
x
x
Home Page
Title Page
y
y
◭◭
◮◮
◭
◮
x
x
Page 8 of 100
None of the above
27. What is the shortest distance between the points (2, −3) and (3, 4)
4 √ 2 2 √ 5 5 √ 2
4 √ 6 3 √ 10
Go Back
Full Screen
Close
Quit

28. x1/2
x 3/5 x −1/10 x 1/10 x 5/6
x 6/5 x 3/10
29.
x 3 y − xy 3
x 2 − 2xy + y 2 =
x + y
xy
xy
x − y
xy
x + y
xy(x + y)
x − y
x + y
x − y
30. A freight train leaves Los Angeles travelling east at a speed of 60 km/h. Two hours later,
a passenger train leaves Los Angeles travelling in the same direction on a parallel track at
90 km/h. How long will it take the passenger train to overtake the freight train
31.
3 hours 3.25 hours 3.5 hours
4 hours 4.5 hours
1
3 √ 2 + 2 equals
3 √ 2 + 2
22
1
5 √ 2
3 √ 2 − 2
14
3 − 2 √ 2
32. Jane has 20 coins in her pocket, each of which is either a quarter or a nickel. If the value
of the coins is $3.20, how many quarters does she have
3 √ 2
20
7 quarters 9 quarters 11 quarters
13 quarters 15 quarters
Home Page
Title Page
◭◭ ◮◮
◭
◮
Page 9 of 100
Go Back
Full Screen
Close
Quit

33. √ 9x 2 + 9y 2
3x + 3y 9xy 9 √ x 2 + y 2
3 √ x 2 + y 2 9x + 9y
34. Solve (x − 1)(x − 5) > 0 for x.
1 < x < 5 x < 1 or x > 5 x < 1
x > 5 5 < x < 1
35. if x + 1 = 1, then x =
x
√
1 3
2 − 2 i √
1 2
2 + 2 i 1 + 2i
2 − i
√ √
2 + 2i
36. What number must be added to x 2 + 7x so that the resulting trinomial is a “perfect square
trinomial”
7
2
49
2
7
14
37. If 3 · 9 x = 1 3 , then x =
49
4
1
2
− 1 2
−2
−1 1
38. Given that y varies directly as x 2 , if y = 9 when x = 2, what is y when x = 3
9/2 81/4 27/4
9/3 81/2
Home Page
Title Page
◭◭ ◮◮
◭
◮
Page 10 of 100
Go Back
Full Screen
Close
Quit

39.
1/x − 1/y
x − y
1
xy
−xy
=
40. log 6 4 + log 6 9 =
− 1
xy
1
xy(x − y)
2 log 6 13 log 6 5
−5 log 6 7
41. If i = √ −1, simplify (1 + √ 3 i) 2 .
−2 4 −2 − √ 3 i
−2 + 2 √ 3 i
2 − 2 √ 3 i
42. The difference of the squares of two consecutive integers is 25. One of the integers is
17 13 11
9 7
43. One factor of x 4 − 5x 2 − 36 is
x + 2 x + 6 x − 9
x + 12 x + 3
44. If x 2 − x − 7 = 0, then x =
3 ± √ 7
2
1 ± √ 29
2
2 ± √ 21
3
−2 ± √ 7
3
xy
1 ± √ 7
2
Home Page
Title Page
◭◭ ◮◮
◭
◮
Page 11 of 100
Go Back
Full Screen
Close
Quit

45. What is the domain of the function f defined by the following equation
f(x) = √ 4 − x
(−∞, 4) (4, ∞) [4, ∞)
(−4, 4) (−∞, 4]
Contents
Home Page
Title Page
◭◭
◮◮
◭
◮
Page 12 of 100
Go Back
Full Screen
Close
Quit

Solutions to Quizzes
Solution to Question 1: First, use the law of exponents (ab) 3 = a 3 b 3 to write
(xy 2 ) 3
x 2 y 5 = x3 (y 2 ) 3
x 2 y 5 .
Next, use the law of exponents (a m ) n = a mn to write
x 3 (y 2 ) 3
x 2 y 5 = x3 y 6
x 2 y 5 = x3
x 2 · y6
y 5 .
Finally, the law of exponents a m /a n = a m−n delivers
Home Page
Title Page
x 3
x 2 · y6
y 5 = xy.
◭◭
◮◮
Hence,
(xy 2 ) 3
x 2 y 5
= xy.
At this point, substitute x = 2 and y = −3 to obtain
xy = (2)(−3) = −6.
Click to return to exam
◭
◮
Page 13 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 2: Factor and cancel.
provided x ≠ 3, −1, or −4.
x + 1
x − 3 · x 2 − x − 6
x 2 + 5x + 4 = x + 1 (x − 3)(x + 2)
·
x − 3 (x + 1)(x + 4)
= x + 2
x + 4 ,
Click to return to exam
Home Page
Title Page
◭◭
◮◮
◭
◮
Page 14 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 3: Multiply both sides of the equation by 5.
x − 4 5 x = 5
5
(x − 4 )
5 x = 5(5)
5x − 4x = 25,
Simplify.
x = 25
Click to return to exam
Home Page
Title Page
◭◭
◮◮
◭
◮
Page 15 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 4: Reorder and multiply.
(2.3 × 10 6 ) × (5.0 × 10 −1 ) = (2.3 × 5.0) × (10 6 × 10 −1 )
= 11.5 × 10 5
Now, 11.5 = 1.15 × 10 1 , and substituting,
11.5 × 10 5 = 1.15 × 10 1 × 10 5
= 1.15 × 10 6 .
Home Page
Click to return to exam
Title Page
◭◭
◮◮
◭
◮
Page 16 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 5: Subtract 1 from both sides of the inequality.
1 − 3x < 7
1 − 3x − 1 < 7 − 1
−3x < 6
Divide both sides of the resulting inequality by −3, reversing the inequality symbol.
−3x
−3 > 6 −3
x > −2
Home Page
Title Page
Click to return to exam
◭◭
◮◮
◭
◮
Page 17 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 6: One technique would involve multiplying numerators and denominators.
a 2 b
35 · 7
b 3 · 1
a = 7a2 b
35ab 3
= 7 35 · a2
a · b
b 3
Cancelling,
Home Page
7
35 · a2
a · b
b 3 = 1 5 · a · 1
b 2
= a
5b 2
Click to return to exam
◭◭
Title Page
◮◮
◭
◮
Page 18 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 7: Since am
a n
= am−n , this problem requires that we subtract exponents.
x −4
x −8 = x−4−(−8) = x −4+8 = x 4
Click to return to exam
Home Page
Title Page
◭◭
◮◮
◭
◮
Page 19 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 8: Substitute x = −2.
x|x| = (−2)| − 2|
= (−2)(2)
= −4
Click to return to exam
Home Page
Title Page
◭◭
◮◮
◭
◮
Page 20 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 9: The slope of a line through (x 1 , y 1 ) and (x 2 , y 2 ) is found with the
formula
m = y 2 − y 1
x 2 − x 1
.
In this case, substitute m = 1/2 and the points A(−3, y) and B(2, 4) to obtain
Cross-multiply,
1
2 = 4 − y
2 − (−3)
1
2 = 4 − y
5
Home Page
Title Page
5 = 2(4 − y)
5 = 8 − 2y
◭◭
◮◮
Subtract 8 from both sides of the equation.
5 − 8 = 8 − 2y − 8
−3 = −2y
◭
◮
Page 21 of 100
Divide both sides of the result by -2.
Hence,
−3
−2 = −2y
−2
3
2 = y
y = 3 2
Go Back
Full Screen
Close
Click to return to exam
Quit

Solution to Question 10: Work the inner parenthesis first, distributing the 2 and simplifying.
2a − [2(3a − b) − 4b] = 2a − [6a − 2b − 4b]
= 2a − [6a − 6b]
Distribute the minus sign and simplify.
2a − [6a − 6b] = 2a − 6a + 6b
= −4a + 6b
Home Page
Click to return to exam
Title Page
◭◭
◮◮
◭
◮
Page 22 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 11: Find a common denominator for the left-hand side and add.
Invert both sides of this result.
1
x + 1 y = 1 z
y
xy + x xy = 1 z
x + y
= 1 xy z
z =
xy
x + y
Home Page
Title Page
Click to return to exam
◭◭
◭
◮◮
◮
Page 23 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 12: Use the law of exponents (xy) n = x n y n to write
(−2ab 2 ) 3 (3ab) = [(−2) 3 a 3 (b 2 ) 3 ](3ab).
Next, use the law of exponents (x m ) n = x mn to write
[
(−2) 3 a 3 (b 2 ) 3] (3ab) = (−8a 3 b 6 )(3ab).
Finally, use the law x m x n = x m+n to simplify.
(−8a 3 b 6 )(3ab) = −24a 4 b 7
Home Page
Title Page
Hence, (−2ab 2 ) 3 (3ab) = −24a 4 b 7
Click to return to exam
◭◭
◮◮
◭
◮
Page 24 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 13: Multiply the first equation by 2.
2x + 2y = 2a
x − 2y = b
Add the resulting equations.
Divide both sides of this result by 3.
3x = 2a + b
Home Page
3x
3 = 2a + b
3
x = 2a + b
3
Click to return to exam
◭◭
Title Page
◮◮
◭
◮
Page 25 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 14: Substitute x = −2 in f(x) = 1 − 2x − x 2 .
f(−2) = 1 − 2(−2) − (−2) 2
Exponents first.
f(−2) = 1 − 2(−2) − 4
Multiplication second.
f(−2) = 1 + 4 − 4
Addition and subtraction last. Thus,
f(−2) = 1.
Click to return to exam
Home Page
Title Page
◭◭
◮◮
◭
◮
Page 26 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 15: Because √ a √ b = √ ab, provided a, b ≥ 0,
√
3
√
21 =
√
3 · 21 =
√
63.
Factor out a perfect square, as in
√
63 =
√
9
√
7 = 3
√
7.
Hence, √
3
√
21 = 3
√
7.
Click to return to exam
Home Page
Title Page
◭◭
◮◮
◭
◮
Page 27 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 16: Because 2x − 5y = 10 is the form Ax + By = C, the graph is a line.
Let x = 0, then
2(0) − 5y = 10
−5y = 10
y = −2
Thus, the y-intercept is (0, −2).
Let y = 0, then
Home Page
Thus, the x-intercept is (5,0). Thus,
2x − 5(0) = 10
2x = 10
x = 5
y
Title Page
◭◭ ◮◮
◭
◮
Page 28 of 100
(5, 0)
x
Go Back
(0, −2)
Full Screen
Close
Click to return to exam
Quit

Solution to Question 17: First, make equivalent fractions with a common denominator.
a
b + c d = a b · d
d + c d · b
b
= ad
bd + bc
bd
Now, add.
Hence,
ad
bd + bc ad + bc
=
bd bd
a
b + c d
ad + bc
= .
bd
Click to return to exam
Home Page
Title Page
◭◭ ◮◮
◭
◮
Page 29 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 18: Use the law of exponents a m /a n = a m−n to write
x 2n+3
x n−2
= x(2n+3)−(n−2)
= x 2n+3−n+2
= x n+5 .
Click to return to exam
Home Page
Title Page
◭◭
◮◮
◭
◮
Page 30 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 19: You should be familiar with the graph of y = x 2 .
y
y = x 2
x
Home Page
Title Page
If we interchange x and y in the equation y = x 2 , we get the equation x = y 2 . Interchanging x
and y like this will reflect the original graph across the line y = x, as shown below.
y
y = x
◭◭
◭
◮◮
◮
y = x 2 x = y 2
Page 31 of 100
Go Back
x
Full Screen
Close
Quit

Hence, the graph of y = x 2 looks as follows.
y
x
Home Page
x = y 2
Title Page
◭◭
◮◮
Click to return to exam
◭
◮
Page 32 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 20: Solve the equation for y. First, subtract 2x from both sides of the
equation.
2x − 5y = 10
2x − 5y − 2x = 10 − 2x
−5y = 10 − 2x
Divide both sides of this result by −5.
−5y
−5
=
10 − 2x
−5
y = 10
−5 − 2x
−5
y = −2 + 2 5 x
Home Page
Title Page
◭◭ ◮◮
Comparing
y = 2 5 x − 2
◭
◮
with
y = mx + b
reveals that m = 2/5 is the slope of the line represented by this equation.
Click to return to exam
Page 33 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 21: By definition,
log a y = x and a x = y
are equivalent. Hence,
log a 2 = 3 is equivalent to a 3 = 2.
Click to return to exam
Home Page
Title Page
◭◭
◮◮
◭
◮
Page 34 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 22: Multiply numerator and denominator of the second fraction by −1.
However, −a + b equals b − a, so
a
b − a +
a
b − a +
b
a − b =
−b
−a + b =
With a common denominator, we can now add.
a
b − a +
−b
−(a − b)
= a
b − a + −b
−a + b
a
b − a +
a
b − a + −b
b − a = a − b
b − a
−b
b − a .
Factor a −1 from the numerator of the fraction on the right hand side of the equation.
Cancel.
Hence,
a − b −(b − a)
=
b − a b − a
−1(b − a)
b − a
= −1
a
b − a + b
a − b = −1.
Click to return to exam
Home Page
Title Page
◭◭ ◮◮
◭
◮
Page 35 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 23: Recall that
and write
a n ( a
) n
b n = b
8 x ( ) x 8
4 x = = 2 x .
4
Click to return to exam
Home Page
Title Page
◭◭
◮◮
◭
◮
Page 36 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 24: Make one side of the equation zero and factor.
x 2 − 7x = 18
x 2 − 7x − 18 = 0
(x − 9)(x + 2) = 0
Because the product is zero, at least one of the factors must equal zero. Thus,
Solving these equations provides the solutions
x − 9 = 0 or x + 2 = 0.
Home Page
Title Page
x = 9 or x = −2.
Click to return to exam
◭◭
◮◮
◭
◮
Page 37 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 25: First,
f(x) − f(2)
x − 2
Factor the numerator and denominator and cancel.
= x2 − 2 2
x − 2 .
x 2 − 2 2
x − 2
=
(x + 2)(x − 2)
x − 2
= x + 2.
Hence,
provided x ≠ 2.
f(x) − f(2)
x − 2
= x + 2,
Click to return to exam
Home Page
Title Page
◭◭
◮◮
◭
◮
Page 38 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 26: Compare y = mx with y = mx + b and note that b = 0. Hence, the
y-intercept is (0,0). Because the slope is m and we are told that m is negative, the line must
go downhill as we sweep our eyes from left to right.
y
Home Page
x
Title Page
◭◭
◮◮
y = mx, m < 0
◭
◮
Click to return to exam
Page 39 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 27: To find the distance between two points (x 1 , y 1 ) and (x 2 , y 2 ), use the
distance formula.
d = √ (x 1 − x 2 ) 2 + (y 1 − y 2 ) 2
Hence, the distance between the points (2, −3) and (3, 4) is
d = √ (2 − 3) 2 + (−3 − 4) 2
= √ (−1) 2 + (−7) 2
= √ 1 + 49
Home Page
Factor out a perfect square.
= √ 50
d = √ 50 = √ 25 √ 2 = 5 √ 2
Click to return to exam
Title Page
◭◭ ◮◮
◭
◮
Page 40 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 28: Because am
a n
= am−n , we subtract exponents in this case.
x 1/2
= x1/2−3/5
x3/5 Get a common denominator, create equivalent fractions, then subtract.
x 1/2−3/5 = x 5/10−6/10
= x −1/10
Home Page
Hence,
x 1/2
= x−1/10
x3/5 Click
to return to exam
Title Page
◭◭ ◮◮
◭
◮
Page 41 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 29: Factor numerator and denominator.
x 3 y − xy 3
x 2 − 2xy + y 2 = xy(x2 − y 2 )
(x − y)(x − y)
xy(x + y)(x − y)
=
(x − y)(x − y)
Cancelling the common factors we get
xy(x + y)(x − y)
(x − y)(x − y)
=
xy(x + y)
.
x − y
Home Page
Title Page
Hence,
provided x ≠ y.
x 3 y − xy 3 xy(x + y)
x 2 = ,
− 2xy + y2 x − y
Click to return to exam
◭◭
◭
◮◮
◮
Page 42 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 30: Let t represent time for passenger train to overtake the freight train.
Because the freight train left 2 hours earlier, its time travelled equals t + 2 hours. Let’s arrange
what we know in tabular form.
Speed (km/h) Time (h) Distance (km)
Passenger 90 t 90t
Freight 60 t + 2 60(t + 2)
When the trains meet, the both have travelled the same distance. Hence,
Home Page
90t = 60(t + 2)
Title Page
Solve for t.
90t = 60t + 120
30t = 120
t = 4
◭◭
◭
◮◮
◮
Hence, it takes 4 hours for the passenger train to overtake the freight train.
Click to return to exam
Page 43 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 31: Multiply numerator and denominator by the conjugate, 3 √ 2 − 2.
1
3 √ 2 − 2 = 3 √ 2 − 2
(3 √ 2 + 2)(3 √ 2 − 2)
The denominator follows the pattern (a + b)(a − b) = (a 2 − b 2 ), so
3 √ 2 − 2
(3 √ 2 + 2)(3 √ 2 − 2 ) = 3 √ 2 − 2
(3 √ 2) 2 − (2) 2
Home Page
= 3√ 2 − 2
9(2) − 4
Title Page
Hence,
= 3√ 2 − 2
18 − 4
= 3√ 2 − 2
14
1
3 √ 2 + 2 = 3√ 2 − 2
14
Click to return to exam
◭◭ ◮◮
◭
◮
Page 44 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 32: Let Q represent the number of quarters and N the number of nickels.
Arrange the given information in tabular form.
Coins Value
Quarters Q 25Q
Nickels N 5N
Totals 20 320
The value in the third column is in cents. Adding the second and third columns,
Q + N = 20
25Q + 5N = 320
Solve the first equation for N by subtracting Q from both sides.
Home Page
Title Page
N =20 − Q
Substitute this result into the second equation.
25Q + 5N = 320
25Q + 5(20 − Q) = 320
◭◭
◭
◮◮
◮
Simplify the left-hand side.
Page 45 of 100
25Q + 100 − 5Q = 320
20Q + 100 = 320
Subtract 100 from both sides, then divide both sides by 20.
20Q + 100 = 320
20Q = 220
Q = 11
Hence, Jane has 11 quarters.
Click to return to exam
Go Back
Full Screen
Close
Quit

Solution to Question 33: We can factor.
√
9x2 + 9y 2 = √ 9(x 2 + y 2 )
= √ 9 √ x 2 + y 2
= 3 √ x 2 + y 2
Note: To go further would be a mistake. Many think that √ x 2 + y 2 = x + y, which is clearly
false as demonstrated by this example.
√
32 + 4 2 = √ 9 + 16 = √ 25 = 5
Home Page
On the other hand,
3 + 4 = 7.
◭◭
Title Page
◮◮
Thus, √
9x2 + 9y 2 = 3 √ x 2 + y 2 .
Click to return to exam
◭
◮
Page 46 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 34: The simplest way to proceed is to understand that
y = (x − 1)(x − 5)
is a parabola opening upward. Solving
0 = (x − 1)(x − 5)
provides x-intercepts at x = 1 and x = 5. Hence the graph of y = (x − 1)(x − 5) follows.
y
Home Page
Title Page
◭◭
◮◮
(1, 0) (5, 0)
x
◭
◮
Page 47 of 100
Go Back
To find the solution of (x − 1)(x − 5) > 0, we note where the graph of the parabola lies above
the x-axis, as noted in the image above. Hence, the solution of (x − 1)(x − 5) > 0 is
x < 1 or x > 5.
Click to return to exam
Full Screen
Close
Quit

Solution to Question 35: Multiply both sides of the equation by x.
Make one side zero.
Recall the quadratic formula. If
then
x + 1 x = 1
(
x x + 1 )
= x(1)
x
( 1
x 2 + x = x
x)
x 2 + 1 = x
x 2 − x + 1 = 0
ax 2 + bx + c = 0,
x = −b ± √ b 2 − 4ac
.
2a
Compare x 2 − x + 1 and ax 2 + bx + c and note that a = 1, b = −1, and c = 1. So,
x = −(−1) ± √ (−1) 2 − 4(1)(1)
2(1)
x = 1 ± √ 1 − 4
2
x = 1 ± √ −3
.
2
Home Page
Title Page
◭◭ ◮◮
◭
◮
Page 48 of 100
Go Back
Full Screen
However, recall that √ −3 = √ 3 i, so
x = 1 2 ± √
3
2 i.
Close
Quit

Hence, the correct “choice” from the set of solutions is
x = 1 2 − √
3
2 i. Click to return to exam
Home Page
Title Page
◭◭
◮◮
◭
◮
Page 49 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 36: The solution is based on the fact that
(a + b) 2 = a 2 + 2ab + b 2 .
Consider x 2 + 7x again. Take half of the coefficient of x and square it, then add the result to
x 2 + 7x, as in
or equivalently,
( ) 2 7
x 2 + 7x + ,
2
( ) 49
x 2 + 7x + .
4
If all is well, this result must be a “perfect square trinomial.” However, it is easy to verify that
(
x + 7 2) 2
= x 2 + 7x + 49 4 .
Home Page
Title Page
◭◭ ◮◮
Therefore, as far as the best choice for this question, 49/4 must be added to x 2 +7x to “complete
the square,” making the resulting trinomial,
◭
◮
a “perfect square trinomial.”
x 2 + 7x + 49
4 ,
Click to return to exam
Page 50 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 37: Put everything in terms of a common base. Since 9 = 3 2 and 1 3 = 3−1 ,
3 · 9 x = 1 3
3 · (3 2 ) x = 3 −1 .
Now, (3 2 ) x = 3 2x , so we can write
3 · 3 2x = 3 −1 .
Next, 3 · 3 2x = 3 1 · 3 2x = 3 1+2x , so we write
3 1+2x = 3 −1 .
Home Page
Title Page
We can now compare exponents.
Solve for x.
1 + 2x = −1
◭◭
◮◮
2x = −2
◭
◮
x = −1
Click to return to exam
Page 51 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 38: To say that y varies directly as x 2 means that
y = kx 2 ,
where k is a constant to be determined. However, we’re given that y = 9 when x = 2. Substitute
these numbers and solve for k.
y = kx 2
9 = k(2) 2
9 = 4k
k = 9 4
Home Page
Title Page
Hence,
y = kx 2
◭◭
◮◮
becomes
y = 9 4 x2 .
◭
◮
Finally, if x = 3, then
y = 9 4 (3)2 = 9 4 (9) = 81 4 .
Page 52 of 100
Click to return to exam
Go Back
Full Screen
Close
Quit

Solution to Question 39: Perhaps the simplest approach is to multiply numerator and denominator
by xy.
1/x − 1/y
x − y
xy(1/x − 1/y)
=
xy(x − y)
xy(1/x) − xy(1/y)
=
xy(x − y)
= y − x
xy(x − y)
Home Page
Factor a −1 from the numerator.
Title Page
Cancel.
y − x −1(x − y)
=
xy(x − y) xy(x − y)
−1(x − y)
xy(x − y) = −1
xy
◭◭
◭
◮◮
◮
Hence,
1/x − 1/y
x − y
= − 1
xy .
Click to return to exam
Page 53 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 40: Recall the law of logarithms.
Using this law,
log a x + log a y = log a xy
log 6 4 + log 6 9 = log 6 (4 · 9)
= log 6 36.
Now, since 6 2 = 36, we know that
log 6 36 = 2.
Click to return to exam
Home Page
Title Page
◭◭
◮◮
◭
◮
Page 54 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 41: Multiply as follows.
(1 + √ 3 i) 2 = (1 + √ 3 i)(1 + √ 3 i)
= 1 + √ 3 i + √ 3 i + 3 i 2
= 1 + 2 √ 3 i + 3 i 2
However, i 2 = −1, so
1 + 2 √ 3 i + 3 i 2 = 1 + 2 √ 3 i − 3.
Home Page
= −2 + 2 √ 3 i.
Title Page
Hence,
(1 + √ 3i) 2 = −2 + 2 √ 3 i.
Click to return to exam
◭◭
◮◮
◭
◮
Page 55 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 42: Let k represent an integer. The next consecutive integer is then k + 1.
The difference of the squares of these integers is 25. Thus, we can write
However, (k + 1) 2 = k 2 + 2k + 1, so
(k + 1) 2 − k 2 = 25.
k 2 + 2k + 1 − k 2 = 25
2k + 1 = 25.
Home Page
Solve for k.
2k = 24
Title Page
k = 12.
◭◭
◮◮
If k = 12, then k + 1 = 13.
Click to return to exam
◭
◮
Page 56 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 43: First,
x 4 − 5x 2 − 36 = (x 2 − 9)(x 2 + 4).
The second factor on the right does not factor, but the first factor is the difference of two
squares. Thus,
x 4 − 5x 2 − 36 = (x + 3)(x − 3)(x 2 + 3).
Note that x + 3 is a factor of x 4 − 5x 2 − 36 and the correct choice for this question.
Click to return to exam
Home Page
Title Page
◭◭
◮◮
◭
◮
Page 57 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 44: Compare
x 2 − x − 7 = 0
with
ax 2 + bx + c = 0,
and note that a = 1, b = −1, and c = −7. Substitute these values into the quadratic formula.
Simplify.
x = −b ± √ b 2 − 4ac
2a
x = −(−1) ± √ (−1) 2 − 4(1)(−7)
2(1)
x = 1 ± √ 1 + 28
2
x = 1 ± √ 29
2
Click to return to exam
Home Page
Title Page
◭◭ ◮◮
◭
◮
Page 58 of 100
Go Back
Full Screen
Close
Quit

Solution to Question 45: The domain of
f(x) = √ 4 − x
is the set of all x that make sense in the equation. Because you cannot take the square root of
a negative number, the expression under the square root must be nonnegative. That is,
4 − x ≥ 0.
Subtract 4 from both sides of this inequality.
4 − x − 4 ≥ 0 − 4
−x ≥ −4
Home Page
Title Page
Multiply both sides of the last inequality by −1, reversing the inequality sign.
◭◭
◮◮
x ≤ 4
There are a number of notations we can use to describe the domain just found. In set-builder
notation, we can write
Domain = {x : x ≤ 4}.
However, the choices are given in interval notation. Note that
(−∞, 4] = {x : x ≤ 4},
◭
◮
Page 59 of 100
Go Back
and is the correct choice for this question.
Click to return to exam
Full Screen
Close
Quit