The document focuses on solving quadratic equations by extracting square roots, targeting grade 9 mathematics students. It outlines key properties of quadratic equations, detailing the number of real solutions based on the value of k (positive, zero, or negative). Various examples illustrate the extraction of square roots and the solving process for different equations.

1. Grade 9 –
Mathematics
Quarter I
SOLVING QUADRATIC EQUATIONS BY
EXTRACTING SQUARE ROOTS

2. Objectives:
1.
familiarize
numbers that
are
perfect squares; and
2. solve quadratic equations by
extracting square roots.

4. Quadratic Equations that can be written in the form 𝒙𝟐 = 𝒌 can be
solved by applying the following properties:
1. If 𝑘 > 0, then 𝒙𝟐 = 𝒌 has two real solutions or roots: 𝑥 = ±
.
𝑘
2. If 𝑘 = 0, then 𝒙𝟐 = 𝒌 has two real solutions or roots: 𝑥 = 0
3. If 𝑘 < 0, then 𝒙𝟐 = 𝒌 has no solutions or roots:
Speaking Mathematically
𝑥 = ± 36 is read as ‘𝒙 is equal to the positive or negative square root of 36’
𝑥 − 5 2is read as “the square root of the square of the quantity 𝒙 𝒎𝒊𝒏𝒖𝒔
𝟓".

5. 𝑥2 = 49
𝑥 = ± 7
How to extract square roots?
√𝑥2=√49

6. 𝑥2 = 169
𝑥 = ± 13
How to extract square roots?
√𝑥2=√169

7. 8 = 4 ∙ 2 32 = 16 ∙ 2 54 = 9 ∙ 6 75 = 25 ∙ 3
12 = 4 ∙ 3 40 = 4 ∙ 10 56 = 4 ∙ 14 76 = 4 ∙ 19
18 = 9 ∙ 2 44 = 4 ∙ 11 54 = 9 ∙ 6 80 = 16 ∙ 5
20 = 4 ∙ 5 45 = 9 ∙ 5 60 = 4 ∙ 15 90 = 9 ∙ 10
24 = 4 ∙ 6 48 = 16 ∙ 3 63 = 9 ∙ 7 96 = 16 ∙ 6
27 = 9 ∙ 3 50 = 25 ∙ 2 68 = 4 ∙ 17 98 = 49 ∙ 2
28 = 4 ∙ 7 52 = 4 ∙ 13 72 = 36 ∙ 2 99 = 9 ∙ 11

8. 8 = 4 ∙ 2 54 = 9 ∙ 6
12 = 4 ∙ 3 56 = 4 ∙ 14
18 = 9 ∙ 2 54 = 9 ∙ 6
20 = 4 ∙ 5 60 = 4 ∙ 15
24 = 4 ∙ 6 63 = 9 ∙ 7
27 = 9 ∙ 3 68 = 4 ∙ 17
28 = 4 ∙ 7 72 = 36 ∙ 2
32 = 16 ∙ 2 75 = 25 ∙ 3
40 = 4 ∙ 10 76 = 4 ∙ 19
44 = 4 ∙ 11 80 = 16 ∙ 5
45 = 9 ∙ 5 90 = 9 ∙ 10
48 = 16 ∙ 3 96 = 16 ∙ 6
50 = 25 ∙ 2 98 = 49 ∙ 2
52 = 4 ∙ 13 99 = 9 ∙ 11
𝑥2 = 75
Get the square root of both sides
Factor the perfect squares
Get the square root of the
perfect square.
𝑥2 = 75
𝑥 = ± 25 ∙
3
𝑥 = ± 25 3
𝑥 = ± 5
3

9. 𝟐 𝒙 − 𝟓 𝟐
= 𝟑𝟐
Divide both sides by 2
Get the square root of both sides
𝒙 − 𝟓 𝟐 = 𝟏𝟔
𝒙 − 𝟓 𝟐 =
𝟏𝟔
𝒙 − 𝟓 = ±
𝟒
𝒙 − 𝟓 = −𝟒
𝒙 = −𝟒 + 𝟓
Find the solutions or roots.
𝒙 − 𝟓
= 𝟒
𝒙 = 𝟒
+ 𝟓
𝟐 𝒙 − 𝟓 𝟐 = 𝟑𝟐
𝟐

10. 𝟑 𝟒𝒙 − 𝟏 𝟐 − 𝟏
= 𝟏𝟏
Divide both sides by 3 𝟑 𝟒𝒙 − 𝟏 𝟐
= 𝟏𝟐
𝟒𝒙 − 𝟏𝟐 = 𝟒
𝟒𝒙 − 𝟏
= 𝟐
4𝒙 = 𝟑
𝟑
𝒙
=
𝟒𝒙 − 𝟏 =
−𝟐
4𝒙 = −𝟏
𝒙
=
−
𝟏
𝟒
Get the square root of both
sides Find the solutions or roots.
𝟒𝒙 − 𝟏 𝟐 =
𝟒
𝟒𝒙 − 𝟏 =
± 𝟐
𝟑 𝟒𝒙 − 𝟏 𝟐 = 𝟏𝟐
𝟑

11. 𝟐𝒙 − 𝟑 𝟐
= 𝟏𝟖
Get the square root of both sides
2𝑥 − 32 = 18
2𝑥 − 3 2 = ± 9
∙ 2
2𝑥 − 3 = ± 9 2
2𝑥 − 3 = ±3 2
2𝑥 − 3 = ±3
2
𝟐𝒙 − 𝟑 = 𝟑
𝟐
𝟐𝒙 = 𝟑 + 𝟑
𝟐
𝒙
=
𝟑 + 𝟑
𝟐
𝟐
𝟐𝒙 − 𝟑 = −𝟑
𝟐
𝟐𝒙 = 𝟑 − 𝟑
𝟐
𝒙
=
𝟑 −𝟑
𝟐
𝟐

12. 𝟐 𝟓𝒙 + 𝟐 𝟐
= 𝟔𝟒
5𝑥 + 2 2 = 32
5𝑥 + 2 2 = ± 16 ∙
2
2
5𝑥 + 2 = ± 16
5𝑥 + 2 = ±4
2
𝒙
=
Divide both sides by 2
5𝑥 + 2 2 = 32
Get the square root of both sides
5𝑥 + 2 = ± 4 2
𝟓𝒙 + 𝟐 = 𝟒𝟐 𝟓𝒙 + 𝟐 = − 𝟒
𝟐
𝟓𝒙 = −𝟐 + 𝟒 𝟐
𝟓𝒙 = −𝟐 − 𝟒 𝟐
𝒙
=
−𝟐 + 𝟒 𝟐 −𝟐 − 𝟒
𝟐
𝟓 𝟓
𝟐 𝟓𝒙 + 𝟐 𝟐
= 𝟔𝟒 �
�