PRACTICE: Mixed practice solving quadratic equations
Use any method you know (factoring, quadratic formula, finding roots, completing
the square) that’s appropriate to solve the quadratic equations below. You may
use your calculator and the formula sheet.
LEVEL 1
1) 7x2 = 63
2) x2 + 5x + 6 = 0
3) x2 + 10x + 21 = 0
4) x2 – 3 = 33
5) 6x2 – 7x – 5 = 0
6) 2x2 + 5x – 12 = 0
D. Stark 3/29/2017
Mixed practice solving quadratic equations 1
LEVEL 2
1) 2x2 = 90
2) x2 – 4 = 16
3) x2 – 8x + 15 = 0
4) x2 + 4x – 32 = 0
5) 2x2 – 11x = 6
6) 15x2 + 2x – 24 = 0
CHALLENGE: x2 – 4x – 6 = 0
D. Stark 3/29/2017
Mixed practice solving quadratic equations 2
PRACTICE: Mixed practice solving quadratic equations
KEY
LEVEL 1
1) 7x2 = 63
2) x2 + 5x + 6 = 0
coefficient of x2 is 1
so will try factoring
no x term so method of roots
7x2 = 63
factored form: (x + )(x + )
factors of 6: 1 & 6; 2 & 3
which sum to 5? 2 & 3
(x + 2)(x + 3) = 0
7x2 = 63
x2 = 9
√𝒙𝟐 = ± √𝟗
x + 2 = 0 or x + 3 = 0
x = –2 or x = –3
solution set: {–3, –2}
x = 3 or –3
solution set: {–3, 3}
For this and other problems, remember
you can plug each solution back into the
original equation to check your answer.
4) x2 – 3 = 33
3) x2 + 10x + 21 = 0
coefficient of x2 is 1
so will try factoring
no x term so method of roots
x2 – 3 = 33
factored form: (x + )(x + )
factors of 21: 1 & 21; 3 & 7
which sum to 10? 3 & 7
(x + 3)(x + 7) = 0
x + 3 = 0 or x + 7 = 0
x = –3 or x = –7
solution set: {–7, –3}
x2 = 36
√𝒙𝟐 = ± √𝟑𝟔
x = 6 or –6
solution set: {–6, 6}
You could also factor a difference
of squares.
x2 – 3 = 33
x2 – 36 = 0
(x + 6)(x – 6) = 0
x + 6 = 0 or x – 6 = 0
x = 6 or –6
solution set: {–6, 6}
When you’ve got perfect squares,
you can use this method instead of the
method of roots if you want.
D. Stark 3/29/2017
Mixed practice solving quadratic equations 3
5) 6x2 – 7x – 5 = 0
6) 2x2 + 5x – 12 = 0
coefficient of x2  1 & lots “–“
so using quadratic formula
−𝒃 ± √𝒃𝟐 − 𝟒𝒂𝒄
𝒙=
b = –7
x=
x=
c = –5
𝒙=
𝟐(𝟔)
𝟕 ± √𝟏𝟔𝟗
𝟏𝟐
𝟕+ 𝟏𝟑
or
𝟏𝟐
𝟐𝟎
=
𝟏𝟐
=
𝟓
𝟑
𝟕 ± 𝟏𝟑
or x =
𝟕 − 𝟏𝟑
x=
𝟏𝟐
−𝟔
𝟏𝟐
𝟏
=–𝟐
𝟏 𝟓
solution set: {– , }
𝟐 𝟑
x=
c = –12
b=5
−𝟓 ± √(𝟓)𝟐 − 𝟒(𝟐)(−𝟏𝟐)
𝟐(𝟐)
𝒙=
𝟏𝟐
x=
𝟐𝒂
a=2
𝟕 ± √(−𝟕)𝟐 − 𝟒(𝟔)(−𝟓)
𝒙=
−𝒃 ± √𝒃𝟐 − 𝟒𝒂𝒄
𝒙=
𝟐𝒂
a=6
𝒙=
coefficient of x2  1
so using quadratic formula
−𝟓 ± √𝟏𝟐𝟏
𝟒
−𝟓 + 𝟏𝟏
𝟒
𝟔
=
𝟒
𝟑
𝟐
=
or
𝟒
x=
or x =
solution set: {–4,
D. Stark 3/29/2017
−𝟓 ± 𝟏𝟏
𝟑
𝟐
−𝟓 − 𝟏𝟏
−𝟏𝟔
𝟒
𝟒
= –4
}
Mixed practice solving quadratic equations 4
LEVEL 2
2) x2 – 4 = 16
1) 2x2 = 90
no x term so method of roots
no x term so method of roots
2x2 = 90
x2 – 4 = 16
2x2 = 90
x2 = 20
x2 = 45
√𝒙𝟐 = ± √𝟐𝟎
√𝒙𝟐 = ± √𝟒𝟓
x = 2√𝟓 or –2√𝟓
solution set: {–2√𝟓, 2√𝟓}
x = 3√𝟓 or –3√𝟓
solution set: {–3√𝟓, 3√𝟓}
3) x2 – 8x + 15 = 0
4) x2 + 4x – 32 = 0
coefficient of x2 is 1
so will try factoring
coefficient of x2 is 1
so will try factoring
factored form: (x – )(x – )
factors of 15: 1 & 15; 3 & 5
how get sum of –8 with both
factors negative? –3 & –5
(x – 3)(x – 5) = 0
factored form: (x + )(x – )
factors of 32: 1 & 32; 2 & 16; 4 & 8
how get sum of 4 with one
factor negative? –4 & 8
(x + 8)(x – 4) = 0
x – 3 = 0 or x – 5 = 0
x = 3 or x = 5
solution set: {3, 5}
x + 8 = 0 or x – 4 = 0
x = –8 or x = 4
solution set: {–8, 4}
since a = 1 and b is even could
also complete the square
since a = 1 and b is even could
also complete the square
x2 – 8x
= –15
2
x – 8x + 16 = –15 + 16
(x – 4)2 = –15 + 16
(x – 4)2 = 1
x2 + 4x
= 32
2
x + 4x + 4 = 32 + 4
(x + 2)2 = 32 + 4
(x + 2)2 = 36
√(𝒙 − 𝟒)𝟐 = ±√𝟏
√(𝒙 + 𝟐)𝟐 = ±√𝟑𝟔
x – 4 = ±1
x – 4 = 1 or x – 4 = –1
x = 5 or x = 3
solution set: {3, 5}
x + 2 = ±6
x + 2 = 6 or x + 2 = –6
x = 4 or x = –8
solution set: {–8, 4}
D. Stark 3/29/2017
Mixed practice solving quadratic equations 5
5) 2x2 – 11x = 6
6) 15x2 + 2x – 24 = 0
Too messy for factoring.
Use quadratic formula.
First, get in standard form
2x2 – 11x – 6 = 0
−𝒃 ± √𝒃𝟐 − 𝟒𝒂𝒄
can try factoring—not too messy
even though a  1
𝒙=
factors of 6: 1 & 6; 2 & 3
factors of 2: 1 & 2 [for the __’s]
what makes FOIL work?
(2x + 1)(x – 6) = 0
2x + 1 = 0 or x – 6 = 0
a = 15
x=–
𝟏
or
𝟐
𝒙=
𝟐𝒂
−𝟐 ± √(𝟐)𝟐 − 𝟒(𝟏𝟓)(−𝟐𝟒)
𝒙=
𝟐(𝟏𝟓)
−𝟐 ± √𝟏𝟒𝟒𝟒
𝟑𝟎
x=6
solution set: {–
𝟏
𝟐
, 6}
x=
x=
c = –24
b=2
−𝟐 + 𝟑𝟖
𝟑𝟎
𝟑𝟔
𝟑𝟎
=
𝟔
𝟓
=
or
x=
or
x=
solution set: {–
−𝟐 ± 𝟑𝟖
𝟑𝟎
−𝟐 − 𝟑𝟖
𝟑𝟎
−𝟒𝟎
𝟒 𝟔
,
𝟑 𝟓
𝟑𝟎
=–
𝟒
𝟑
}
CHALLENGE: x2 – 4x – 6 = 0
coefficient of x2 is 1 so will try factoring
factored form: (x – )(x + )
factors of 6: 1 & 6; 2 & 3
how get sum of –4 with one factor negative? can’t do it—not factorable
could use quadratic formula:
a=1
𝒙=
𝒙=
b = –4
𝒙=
−𝒃 ± √𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
c = –6
𝟒 ± √(−𝟒)𝟐 − 𝟒(𝟏)(−𝟔)
𝟐(𝟏)
𝟒 ± √𝟒𝟎
𝟐
=
𝟒 ± 𝟐√𝟏𝟎
𝟐
=
𝟒
𝟐
±
𝟐√𝟏𝟎
𝟐
[split the fractions to simplify]
x = 2 ± √𝟏𝟎
x = 2 + √𝟏𝟎
or
x = 2 – √𝟏𝟎
D. Stark 3/29/2017
solution set: {2 – √𝟏𝟎, 2 + √𝟏𝟎}
Mixed practice solving quadratic equations 6
or you could complete the square
x2 – 4x = 6
x2 – 4x + 4 = 6 + 4
(x – 2)2 = 6 + 4
(x – 2)2 = 10
√(𝒙 – 𝟐)𝟐 = ±√𝟏𝟎
x – 2 = ±√𝟏𝟎
x – 2 = √𝟏𝟎
or
x – 2 = –√𝟏𝟎
x = 2 + √𝟏𝟎
or
x = 2 – √𝟏𝟎
solution set: {2 – √𝟏𝟎, 2 + √𝟏𝟎}
D. Stark 3/29/2017
Mixed practice solving quadratic equations 7