Illustration of quadratic equations

1. MATHEMATICS IX
Quarter 1, week 1
Quadratic Equations
REYNALDO M. ANARIO, JR.
SubjectTeacher

2. Lesson 1
Objective: The Learner illustrates quadratic
equations
What is Quadratic Equation in
one variable?
Quadratic Equation in one variable is a mathematical
sentence of degree 2 that can be written in the
following standard form,
𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
Where a, b, and c are real numbers and a≠0

3. Examples:
x2 – 4 = 0
y2 = 1
3a2 + 7a - 6 = 0
t + t2 = 6
25 = s2
32 =
r2
3x(x – 2) =
10
When simplify
3x2 – 6x =
10
3x2 – 6x – 10 = 10 –
10
3x2 – 6x – 10 =
0
Standard
form

4. Standard
Form ax2 + bx + c =
0
ax
2
bx c
Quadrati
c
Term
Linear
Term
Constant
Term
x2 + 7x - 6 = 0
x
2
7x -
6
Quadrati
c
Term
Linear
Term
Constant
Term
a = 1 b = 7 c = -6

5. The equation (2x + 5)(x – 1) = -6 is also a
quadratic equation but is not written standard
form.
Transform into standard
form
(2x + 5)(x – 1) =
-6
2x2 – 2x + 5x – 5 =
-6
2x2 + 3x – 5 =
-6
2x2 + 3x – 5 + 6 = -6
+ 6
2x2 + 3x + 1 =
0
Standard
Form
2x
2
3x
1
Quadrati
c
Term
Linear
Term
Constant
Term
a = 2 b = 3 c = 1

6. The length of a rectangular lot is 15m longer than
its width and the area is 100m2.
Area of a
rectangle
A = Length x
Width
width =
y
length = y +
15
area = 100
100 = y(y +
15)
100 = y2 + 15y
y2 + 15y = 100
y2 + 15y – 100 = 100 –
100
y2 + 15y – 100 =
0

7. Lesson 2
Objective: The Learner solves quadratic
equation by: (a) extracting the square
roots; (b) factoring; (c) completing the
square; (d) using the quadratic formula
How to solve quadratic
equation?
Solving Quadratic Equation is also finding the solution/s
of the given quadratic equation.
Solution/s of a quadratic equation is/are represented by
the value of its variable.This value is called the
root/s of a quadratic equation.

8. Solving Quadratic
Equation
b
y
Extracting Square
Root
Factoring
Completing the
Square
Quadratic Formula

9. Solving Quadratic
Equation
b
y
Extracting Square
Root
Perfect
Squares
Roots
𝟒𝟗
− 𝟏𝟎𝟎
𝒙𝟐
𝒙𝟒
(𝒙 + 𝟐)𝟐
±𝟕
−𝟏𝟎
𝒙
𝒙𝟐
𝒙 + 𝟐

10. Extracting Square
Root
REMEMBER 𝒙𝟐
= 𝒌, where k is the
constant
If k > 0, then 𝑥2
= 𝑘 has two real solutions or
roots; 𝑥 = ± 𝑘
If k = 0, then 𝑥2
= 𝑘 has one real solution or root;
𝑥 = 0
If k < 0, then 𝑥2
= 𝑘 has no real solutions or

11. Extracting Square
Root
Example:
Find the solutions of the equation 𝑥2
− 16 = 0
by extracting square roots.
𝒙𝟐
= 𝒌
𝒙𝟐
− 𝟏𝟔 = 𝟎
𝒙𝟐
− 𝟏𝟔 + 𝟏𝟔 = 𝟎 + 𝟏𝟔
𝒙𝟐
= 𝟏𝟔
𝒙𝟐 = ± 𝟏𝟔
𝒙 = ±𝟒
𝒙𝟏 = 𝟒 , 𝒙𝟐 = −𝟒

12. Extracting Square
Root
To check:
Substitute these values (roots) in the original equation.
𝒙𝟐
− 𝟏𝟔 = 𝟎
𝒙𝟏 = 𝟒
𝒙𝟐
− 𝟏𝟔 = 𝟎
𝟒𝟐
− 𝟏𝟔 = 𝟎
𝟏𝟔 − 𝟏𝟔 = 𝟎
𝟎 = 𝟎
𝒙𝟐 = −𝟒
𝒙𝟐
− 𝟏𝟔 = 𝟎
(−𝟒)𝟐
−𝟏𝟔 = 𝟎
𝟏𝟔 − 𝟏𝟔 = 𝟎
𝟎 = 𝟎

13. Extracting Square
Root
Example:
Find the solutions of the equation 𝑥 − 4 2
− 25 =
0. 𝒙𝟐
= 𝒌
𝒙 − 𝟒 𝟐
− 𝟐𝟓 = 𝟎
𝒙 − 𝟒 𝟐
− 𝟐𝟓 + 𝟐𝟓 = 𝟎 + 𝟐𝟓
𝒙 − 𝟒 𝟐
= 𝟐𝟓
𝒙 − 𝟒 𝟐 = ± 𝟐𝟓
𝒙 − 𝟒 = ±𝟓
𝒙𝟏 = 𝟓 + 𝟒
𝒙 − 𝟒 + 𝟒 = ±𝟓 + 𝟒
𝒙 = ±𝟓 + 𝟒
𝒙𝟐 = −𝟓 + 𝟒
𝒙𝟏 = 𝟗 𝒙𝟐 = −𝟏

14. Extracting Square
Root
To check:
Substitute these values (roots) in the original equation.
𝒙𝟏 = 𝟗
(𝒙 − 𝟒)𝟐
−𝟐𝟓 = 𝟎
(𝟗 − 𝟒)𝟐
−𝟐𝟓 = 𝟎
(𝟓)𝟐
−𝟐𝟓 = 𝟎
𝟐𝟓 − 𝟐𝟓 = 𝟎
𝟎 = 𝟎
𝒙 − 𝟒 𝟐
− 𝟐𝟓 = 𝟎
𝒙𝟐 = −𝟏
(𝒙 − 𝟒)𝟐
−𝟐𝟓 = 𝟎
(−𝟏 − 𝟒)𝟐
−𝟐𝟓 = 𝟎
(−𝟓)𝟐
−𝟐𝟓 = 𝟎
𝟐𝟓 − 𝟐𝟓 = 𝟎
𝟎 = 𝟎

15. Solving Quadratic
Equation
b
y
Factoring
Factorable
expressions
Factor/s
𝟐𝒙𝟐
− 𝟖𝒙
−𝟑𝒙𝟐
+ 𝟗𝒙
𝒙𝟐
+ 𝟖𝒙 + 𝟏𝟐
𝟐𝒙(𝒙 − 𝟒)
−𝟑𝒙(𝒙 − 𝟑
(𝒙 + 𝟔)(𝒙 + 𝟐)
𝒙𝟐
+ 𝟓𝒙 − 𝟔 (𝒙 + 𝟔)(𝒙 − 𝟏)
𝟐𝒙𝟐
− 𝟗𝒙 + 𝟕 (𝟐𝒙 − 𝟕)(𝒙 − 𝟏)

16. Factoring
Step
s: 1. Transform the quadratic equation into standard
form if necessary.
2. Factor the quadratic expression.
3. Apply the zero product property by setting each
factor of the quadratic expression equal to 0.
4. Solve each resulting equation.
5. Check the values of the variable obtained by
substituting each in the original equation.

17. Factoring
Examp
le Find the solutions of the equation 𝑥2
+ 8𝑥 = −15.
𝒙𝟐
+ 𝟖𝒙 = −𝟏𝟓
𝒙𝟐
+ 𝟖𝒙 + 𝟏𝟓 = −𝟏𝟓 + 𝟏𝟓
𝒙𝟐
+ 𝟖𝒙 + 𝟏𝟓 = 𝟎
(𝒙 + 𝟓)(𝒙 + 𝟑) = 𝟎
𝒙 + 𝟓 = 𝟎 𝒙 + 𝟑 = 𝟎
𝒙 + 𝟓 − 𝟓 = 𝟎 − 𝟓
𝒙𝟏 = −𝟓
𝒙 + 𝟑 − 𝟑 = 𝟎 − 𝟑
𝒙𝟐 = −𝟑

18. Factoring
To check:
Substitute these values (roots) in the original equation.
𝒙𝟏 = −𝟓
(−𝟓)𝟐
+𝟖 −𝟓 = −𝟏𝟓
𝟐𝟓 − 𝟒𝟎 = −𝟏𝟓
−𝟏𝟓 = −𝟏𝟓
𝒙𝟐
+ 𝟖𝒙 = −𝟏𝟓
𝒙𝟏 = −𝟑
(−𝟑)𝟐
+𝟖 −𝟑 = −𝟏𝟓
𝟗 − 𝟐𝟒 = −𝟏𝟓
−𝟏𝟓 = −𝟏𝟓

19. Factoring
Examp
le The length of a rectangular lot is 15m longer than
its width and the area is 50m2. Find its
dimensions.
Area of a
rectangle
A = Length x
Width
width =
y
length = y +
15
area =
100 100 = y(y +
100 = y2 +
15y
y2 + 15y =
100
y2 + 15y – 100 = 100 –
100
y2 + 15y – 100 =

20. Factoring
Examp
le The length of a rectangular lot is 15m longer than
its width and the area is 100m2. Find its
dimensions.
𝑦2 + 15𝑦 – 100 = 0
𝑦 + 20 𝑦 − 5 = 0
𝑦 + 20 = 0 𝑦 − 5 = 0
𝑦 + 20 − 20 = 0 − 20
𝑦1 = −20
𝑦 − 5 + 5 = 0 + 5
𝑦1 = 15
width= 𝑦
length= 𝑦 + 5
length= 15 + 5
length= 20𝑚
width= 15𝑚

21. Thank you for
watching!