COMPASS
Practice Test 13
Quadratics
Quadratics
• This slide presentation will focus on
quadratics.
• Quadratics will always have a variable
raised to the second power, like x2.
• Factoring is a skill that will help you find
solutions to quadratic equations.
0 = x2 - 6x - 16
0 = (x - 8)(x + 2)
x = {-2, 8}
Quadratic
Formula
 b  b  4ac
x
2a
2
• If you do not like to factor you can always
use the quadratic formula.
0 = x2 - 6x - 16
2
 (6)  (6)  4(1)( 16)
x
a=1
2(1)
b = -6
6

10
6

100
6

36

64
c = -16 x 


2
2
2
6  10 16
6  10  4
x

{
8
,

2
}
x

x

 2
8
2
2
2
2
1. If x = -1 and y = -2, what is the
value of the expression 2x2y- 3xy ?
 A.
 B.
 C.
 D.
 E.
-24
-10
-2
2
10
2 y – 3xy
2x
We start this practice with a
substitution
problem,
not a
= 2(-1)2 (-2)
– 3(-1)(-2)
quadratic.
COMPASS
often
= 2(1) (-2)- 3(-1)(-2)
starts with a substitution
=
-4
–
6
=
-10
problem.
Answer B
2. What are the solutions to the quadratic
x2 - 2x - 48 = 0?
 A.
 B.
 C.
 D.
 E.
6 and 8
-6 and -8
-6 and 8
6 and -8
3 and 16
x2 - 2x - 48 = 0
(x - 8)(x + 6) = 0
Set each factor to 0
x-8=0
x=8
x+6=0
x = -6
x = { 8, -6}
2. What are the solutions to the quadratic
x2 - 2x - 48 = 0?
 A.
 B.
 C.
 D.
 E.
6 and 8
-6 and -8
-6 and 8
6 and -8
3 and 16
2  14 16
8
x

2
2
Or you could find the answer
with the quadratic formula.
a = 1 b = -2 c = 48
 (2)  (2) 2  4(1)( 48)
x
2(1)
2  14
2  4  192 2  196



2
2
2
2  14  12
x

 6
2
2
x  {8,6}
2. What are the solutions to the quadratic
x2 - 2x - 48 = 0?
 A.
 B.
 C.
 D.
 E.
6 and 8
-6 and -8
-6 and 8
6 and -8
3 and 16
(6Another
) 2  2(6) way
48 to
0 find the
solution
to check
each of
36  12is
 48
0
the answers
24  48 back
 0 into the
original equation.
 24  0
This would take a long time,
False
but remember this test is
Thus wenotcan
eliminate
timed.
answers
A6and D
Try x =
This process of elimination is a good strategy if you get stuck.
3. What is the sum of the solutions to the
quadratic x2 - 2x - 48 = 0?
 A.
 B.
 C.
 D.
 E.
14
-14
2
-2
19
To prevent people from using
the process of elimination
discussed on the previous
slide the questions are
sometimes written this way.
Find the solution set {-6, 8}
Add the solutions -6 + 8 = 2
4. What is the sum of the solutions of the
quadratic equation x2 + 3x = 28?
 A.
 B.
 C.
 D.
 E.
3
-3
11
-11
10
First write the equation in standard form.
x2 + 3x - 28 = 0
Using the quadratic formula.
a = 1 b = 3 c = -28
 (3)  (3) 2  4(1)( 28)  3  9  112
x

2(1)
2
 3  121  3  11


x  {4,7}
2
2
 3  11 8
 3  11  14
x
 4 x

 7 4  (7)  3
2
2
2
2
5. What is the sum of the solutions of the
quadratic equation 2x2 - x = 15?
 A. 1 First write the equation in standard form.
 B.
 C.
 D.
 E.
2
2x2 - x - 15 = 0
1
Using the quadratic formula.

2
a
=
2
b
=
-1
c
=
-15
11

2

(

1
)

(

1
)
 4(2)( 15) 1  1  120
2x

4
2(2)
11
1 11
12
1 121

x 3

2
4
4
4
-1 x  1  11   10   5
5 1
3

4
4
2
2 2
6. If the equation x2 - x = 6 is solved for x,
what is the sum of the solutions?
 A. 3
 B. 2
 C. 5
 D. 1
 E. -1
First write the equation in standard form.
x2 - x - 6 = 0
Using the quadratic formula.
a = 1 b = -1 c = -6
 (1)  (1) 2  4(1)( 6)
1  1  24
x

2
2(1)
1 25 1 5 x  1  5  6  3


2
2
2
2
1 5  4
x

 2
3  2  1
2
2
7. What are the solutions to the quadratic
x2 - 5x = -6?
 A.
 B.
 C.
 D.
 E.
-2, -3
2, 3
1, 6
-1, -6
-2, 3
First write the equation in standard form.
x2 - 5x + 6 = 0
Using the quadratic formula.
a = 1 b = -5 c = 6
 (5)  (5) 2  4(1)(6) 5  25  24
x

2(1)
2
5 1
5 1


2
2
5 1 6
5 1 4
x
 3
x
 2
2
2
2
2
x  {2, 3}
8. For all x ≠ 2,
 A. (x + 5)
 B. (x - 2)
 C. (x + 2)
 D. (x - 3)
 E. (x + 3)
x  5x  6
?
x2
2
Factor the numerator.
( x  2)( x  3)
 ( x  3)
( x  2)
8. For all x ≠ 2,
x  5x  6
?
x2
2
 A. (x + 5) Another way to work this problem
 B. (x - 2)
 C. (x + 2)
 D. (x - 3)
 E. (x + 3)
is to just make up a number for x.
Let x = 5
(5)  5(5)  6 6
 2
3
52
2
Now plug x = 5 into each of the
answers until you find a match.
D  ( x  3)  5  3  2
9. If x = -4 is a solution to the equation
x2 + 11x + K = 0, then K = ?
 A. 16
 B. 28
 C. -28
 D. 60
 E. -60
First substitute x = -4 into the given
equation. Then solve for K.
x2 + 11x + K = 0
(4)  11(4)  K  0
16  44  K  0
 28  K  0
K  28
2
10. What are the solutions to the
quadratic x2 - 10x + 24 = 0?
x2 - 10x + 24 = 0
 A. 4 and 6
(x
4)(x
6)
=
0
 B. -4 and 6
x
4
=
0
 C. -4 and -6
x
=
4
 D. 2 and -12
x
6
=
0
 E. -2 and 12
x=6
x = { 4, 6}