1. Practice Test
Unit 2 (Part 2)

2. x2 – 6x + 8 = 0 (x – 2)(x – 4) = 0 , x – 4 = 0 x – 2 = 0 x = 2 x = 4
If x2 – 6x + 8 = 0 and y = x + 3, then what are the possible values of y ? 1
x2 – 6x + 8 = 0
(x – 2)(x – 4) = 0
, x – 4 = 0
x – 2 = 0
x = 2
x = 4
Let x = 2:
y = x + 3
Let x = 4:
y = x + 3
= 2 + 3
= 4 + 3
= 5
= 7

3. 2
(2x + 3y)2 – (2x – 3y)2
(2x + 3y)(2x + 3y) – (2x – 3y) (2x – 3y)
(4x2 + 6xy + 6xy + 9y2) – (4x2 – 6xy – 6xy + 9y2)
(4x2 + 12xy + 9y2) – (4x2 – 12xy + 9y2)
4x2 + 12xy + 9y2 – 4x2 + 12xy – 9y2
24xy

4. (x – y)2 = (x – y)(x – y) = x2 – xy – xy + y2 = x2 – 2xy + y23
If x2 + y2 = 37 and xy = 24, what is the value of (x – y)2?
(x – y)2
= (x – y)(x – y)
= x2 – xy – xy + y2
x2 + y2 = 37
= x2 – 2xy + y2
= x2 + y2 – 2xy
= – 2xy
xy = 24
= – 2(24)
= – 48
= –11

5. If (–8x + 3)(–4x2 + 4x + 6) = ax3 + bx2 + cx + d for all real values of x, what is the value of c ?18
32x3 – 44x2 – 36x + 18
(Add matching colors)

6. (y – 5)2 = 0 Find y2 – 2y when y = 5 (y – 5)(y – 5) = 0 (5)2 – 2(5)
If (y – 5)2 = 0, then find the value of y2 – 2y ?
(y – 5)2 = 0
Find y2 – 2y
when y = 5
(y – 5)(y – 5) = 0
(5)2 – 2(5)
y – 5 = 0 ,
y – 5 = 0
25 – 10
y = 5
y = 5 15

7. 6
If x and y are positive integers, then which of the following must be equal to ?

8. 7
Step 1
Step 2
Step 3

9. 8
If and ab  0, then
1  a3 = b  x
Reciprocal
a3 = bx

10. 9 LCD = 4m2 m2 – 8m + 12 = 0 (m – 2)(m – 6) = 0 m2 + 12 = 8m
Solve for m. 9
LCD = 4m2
m2 – 8m + 12 = 0
(m – 2)(m – 6) = 0
m = 8m
m – 2 = 0 ,
m – 6 = 0
m2 – 8m = 0
m = 2
m = 6

11. 7 + 4 11 Find x + 4 when x = 7 4x – 3 = 25 +3 +3 4x = 28 x = 7
If , then find
the value of x + 4 ? 10
Find x + 4
when x = 7
7 + 4
4x – 3 = 25 +3 +3 11
4x = 28
x = 7

12. 49 < x – 2 < 64 51 < x < 66 +2 +2 +2
If x is an integer and , how
many different values of x are possible? 11
First, solve
inequality.
49 < x – 2 < 64 +2 +2 +2
51 < x < 66

13. 51 < x < 66 If x is an integer and , how 11
many different values of x are possible? 11
51 < x < 66
Integers between
51 and 66 OR
66 – 51 = 15
15 – 1 = 14
Answer: 14 Values

14. 11 Try this strategy How many values of m are possible?
Use an inequality with two numbers
that are close together.
2 < m < 7
Integers between 2 and 7
How many values
of m are possible?
10 < m < 45
Answer: 4 Values OR
45 – 10 = 35
35 – 1 = 34
7 – 2 = 5
5 – 1 = 4

15. 12
–16
–16
w = 81

16. If and , then what is the
value of b ? 13
a = 64
b = 16

17. 14
–3x + 13 < –14
–13
–13
–3x < –27
x > 9

18. 10 < x x 10 < x x –10 10 < –10 –5 10 < –5 –1 < –10
If , then which of the following values could be x ? 15
Strategy: Test each answer by substituting for x.
A. –10
B. –5 10
< x x 10
< x x
–10 10
< –10 –5 10
< –5
–1 < –10
–2 < –5 NO NO

19. 10 < x x 10 < x x 10 < x x 1 10 < 1 2 10 < 2 5 10
If , then which of the following values could be x ? 15
C. 1
D. 2
E. 5 10
< x x 10
< x x 10
< x x 1 10
< 1 2 10
< 2 5 10
< 5
10 < 1
5 < 2
2 < 5 NO NO
YES

20. 16 2x = 1 NO NO If , then which of the
following statements must be true? 16
Strategy: Substitute ¼ for x in each answer.
2x = 1 NO NO

21. 16 2x > 1 NO NO If , then which of the
following statements must be true? 16
Strategy: Substitute ¼ for x in each answer.
2x > 1 NO NO

22. z < 0 17  y = ( + ) positive y > 0 x > y 
If x > y and y > 0 and xz < 0, then which
of the following must be true about all the values of z? 17 
y = ( + ) positive
y > 0
x > y 
x = ( + ) positive
xz < 0 
xz = ( – ) negative
(+)z = ( – )
z < 0
(+)(–) = ( – )
z is negative

23. 18
If the sum of two integers x and k is less than x, which of the following must be true?
x + k < x –x –x
k < 0

24. Which of the following equations represents the statement above? 19
Twice the difference between a certain number and its square root is 15 more than twice the number.
Which of the following equations represents the statement above? 19

25. If a number is doubled and then increased by 10, the result is 5 less than the square of the number.
Which of the following equations represents the statement above? 20
2N + 10 = N2 – 5
2N + 10 = N2 + 5

26. x – 2 3(x – 2) 3(x – 2) = x + 6 2x – 6 = 6 3x – 6 = x + 6 +6 +6 –x –x
If 2 is subtracted from a number and this difference is tripled, the result is 6 more than the number. Find the number. 21
Let number = x
‘2 is subtracted from a number’ =
x – 2
3(x – 2)
‘The difference is tripled’ =
3(x – 2) = x + 6
2x – 6 = 6
3x – 6 = x + 6 +6 +6 –x –x
2x = 12
2x – 6 =
x = 6

27. Let 1st integer = x 13 Let 2nd integer = x + 2 13+2 = 15
If the sum of two consecutive odd integers is 28, what is the product? 22
Let 1st integer = x 13
Let 2nd integer = x + 2
13+2 = 15
x + x + 2 = 28
Sum =
= 28
2x = 28 –2 –2
2x = 26
Product
= 13  15
= 195
x = 13

28. –15 3 –5 –2 Let integer = x 2x + 10 = x2 – 5 –2x – 10 –2x – 10
If a positive integer is doubled and then increased by 10, the result is 5 less than the square of the integer. What is the integer? 23
Let integer = x
2x = x2 – 5
–15
–2x – 10
–2x – 10 3 –5
0 = x2 – 2x – 15 –2
0 = (x + 3)(x – 5)
Two numbers with
Product of –15 and
Sum of –2
x + 3 = 0 ,
x – 5 = 0
x = –3
x = 5

29. Cost of pencil = A Cost of pen = B Jon  1A + 2B = 3.50 Lauren 
Jon buys one pencil and two pens for $ Lauren buys four pencils and three pens for $ How much would one pencil and one pen cost? 24
Cost of pencil = A
Cost of pen = B
Jon 
1A + 2B = 3.50
(Multiply Jon by –4)
Lauren 
4A + 3B = 5.50
Jon 
–4A – 8B = –14.00
Add
Equations
Lauren 
4A + 3B =
– 5B = –8.50
B =
(Pen cost)

30. the original equations. 1A + 2B = 3.50 1A + 2B = 3.50 4A + 3B = 5.50
Jon buys one pencil and two pens for $ Lauren buys four pencils and three pens for $ How much would one pencil and one pen cost? 24
Cost of pencil = A
Cost of pen = B
B = 1.70 (Pen cost)
Find A. Use one of
the original equations.
1A + 2B = 3.50
1A + 2B = 3.50
4A + 3B = 5.50
A + 2(1.70) = 3.50
Cost of pencil and pen
= A B
= = 1.80
A = 3.50
A = 0.10

31. If y varies directly as x2, and y = 3 when
x = 3, what is the value of y when x is 6? 25
y varies directly as x
y varies directly as x2

32. 5  p = 2  100 5p = 200 p = 40 26 Direct Variation
Students receive 5 bonus points for every 2 community service projects they perform. If Mark received 100 bonus points, how many projects did he perform? 26
Note: As the bonus points increase, the community service projects should increase.
Direct
Variation
5  p = 2  100
5p = 200
p = 40

33. 27
If it takes 4 men 3 hours each to pave a playground, how many hours will it take
12 men to complete the same task?
Note: Increasing the number of men will decrease
the amount of time to complete the task.
Inverse
Variation
M1H1 = M2H2
4 · 3 = 12 · H2
x1y1 = x2y2
12 = 12H2
1 Hour
1 = H2

34. f(3) = 33 + 3(3) + 30 f(3) = 27 + 9 + 1 f(3) = 37 What is the value of28
What is the value of
f(x) = 3x + 3x + 30 if x = 3 ?
f(3) = (3) + 30
f(3) =
f(3) = 37

35. 29
If f(x) = x + 2x, what is the
value of f(–2)?
f(–2) = –2 + 2–2

36. x – 5 > 0 +5 +5 x > 5 Find the domain for30
Find the domain for
Note: We can only evaluate the square root
of numbers greater than or equal to zero.
Let expression inside radical be > 0.
x – 5 > 0 +5 +5
x > 5

37. 31
The amount a restaurant owner pays for coffee beans is directly proportional to the number of pounds of coffee she buys. If she buys n pounds of coffee at d dollars per pound, what is the total amount she pays, in dollars, in terms of n and d.
n lb.
d dollars/lb.
total
1 lb.
$3 dollars/lb.
1  3 = $3
2 lb.
$3 dollars/lb.
2  3 = $6
3 lb.
$3 dollars/lb.
3  3 = $9
n lb.
$d dollars/lb.
n  d

38. 32
The cost of preparing for a book sale is $30. If each book is sold for $3.00, express the profit as a function of n, where n represents the number of books sold.
Book
Cost
# of books
sold
Preparation
Cost
Profit 3 1 30
3(1) – 30
= 3–30
= –27 3 2 30
3(2) – 30
= 6–30
= –24 3 3 30
3(3) – 30
= 9–30
= –21 3 4 30
3(4) – 30
= 12–30
= –18

39. 32
The cost of preparing for a book sale is $30. If each book is sold for $3.00, express the profit as a function of n, where n represents the number of books sold.
Book
Cost
# of books
sold
Preparation
Cost
Profit 3 6 30
3(6) – 30
= 18–30
= –12 3 10 30
3(10) – 30
= 30–30
= 0 3 12 30
3(12) – 30
= 36–30
= 6 3 n 30
3n – 30
f(n) = 3n – 30

40. 33
Morgan’s plant grew from 42 centimeters to 57 centimeters in a year. Linda’s plant, which was 59 centimeters at the beginning of the year, grew twice as many centimeters as Morgan’s plant did during the same year. How tall, in centimeters, was Linda’s plant at the end of the year?
Step 1
Centimeters Morgan’s plant grew = 57 – 42 = 15 cm.
Step 2
Twice centimeters Morgan’s plant grew = 2(15) = 30 cm.
Step 3
Height of Linda’s plant = 59 cm + Step 2
= 59 cm + 30 cm = 89 cm.

41. 34
If , then h(x) is
Always Positive
Never Negative
III. Always an Integer
False
False
Note 1
We can only evaluate the square root
of numbers greater than or equal to zero.
Note 2
Note 3

42. (31) = 31 + 12 = 3 + 1 = 4 (31)1 = 41 = 41 + 12 = 4 + 1 = 5 35
For all numbers x and y, let xy be defined as xy = xy + y2. What is the value of (31)1 ?
(31)
= 31 + 12 = =
(31)1
= 41
= 41 + 12 = =

43. {–5,5} x2 – 25 = 0 (x – 5)(x + 5) = 0 x – 5 = 0 , x + 5 = 0 +5 +5 –536
Which values are not in the domain of
Let denominator = 0
Solve equation for x.
x2 – 25 = 0
(x – 5)(x + 5) = 0
x – 5 = 0 ,
x + 5 = 0 +5 +5 –5 –5
{–5,5}
x = 5
x = –5

44. 37
The sign-up fee at a gym is $50. Members then must pay $25 each month. Express the cost of using the gym as a function of m, where m represents the number of months the member participates.
Sign-up
Fee
# of
Months
Monthly
Cost
Gym
Cost 50 1
25(1)
(1)
= 75 50 2
25(2)
(2)
= 100 50 3
25(3)
(3)
= 125 50 m
25(m)
(m)
f(m) = m

45. 38 # of Rings Total rings left 1 7(1) = 7 23 – 7 = 16 NO # of Rings
Each time Shannon pushes the button on a machine, a bell rings 7 times. Each time she turns the switch on the machine, the bell rings 3 times. During one hour, Shannon caused the bell on the machine to ring 23 times. How many times did she push the button?
# of Rings
Total rings left
Pushed Button 1
7(1) = 7
23 – 7 = 16 NO
There are 3 rings each time a switch is turned.
Thus, the number of rings must be a multiple of 3.
# of Rings
Total rings left
Pushed Button 2
7(2) = 14
23 – 14 = 9
YES
Turned Switch 3
3(3) = 9

46. 39 Which system of inequalities best represents the graph? A. Wrong B.D.
The dotted line has a negative slope.
The 2nd inequality for each answer has a negative slope.
The inequality sign should be < or > .

47. 39 Which system of inequalities best represents the graph? B. C. D.
Wrong
The solid line has a positive slope.
The 1st inequality for each answer has a positive slope.
Inequality sign should be < . The shading is below the line.

48.   39 Which system of inequalities best represents the graph? B. C. 32
- Answers B and C have
different slopes for
the solid line. 
- Find two points on the line.
- Use to get to each point on the line.
This will determine the slope and equation of the line.

49. y < –x – 1 –2x + y > –2 y > 2x – 2 40 Negative Slope
Determine the solution of the system of inequalities.
y < –x – 1
Negative Slope
(Shaded Below)
–2x + y > –2
y > 2x – 2
Positive Slope
(Shaded Above) A. B. C. D.