NARCISO SCHOOL INC.

4-4 Maya St. Sta. Maria Village, Balibago, Angeles City, Pampanga
Contact Number: 0919-672-9128/0995-055-8114
Name: Vaughn Matthew De Dios Arcala
Grade 9
Teacher: Ms. Daniela A. Ponce Patterns and Algebra

Mathematics 9
First Quarter
Quadratic “Quadratic Equations”
(Module 1 – Module 2)
Equations
Definition 1.1

A quadratic equation is a second degree equation, which can be

expressed in the standard form a x 2+ bx+ c=0, where a, b, and c are real numbers
and a ≠ 0. The first term, a x 2 is called the quadratic term, bx is the linear term and
c is the constant term.

quadratic term
constant term

linear term
The solutions of a quadratic equation are also called the roots of the equation.
The roots of the quadratic equation a x 2+ bx+ c=0 are zeroes or x-intercepts of the
associated quadratic function f ( x )=a x 2 +bx +c .

a x 2+ b x+ c=0
The Standard Form of
Quadratic Equations
A quadratic equation is an equation that can be written in the form
Where:
 a, b, and c are real numbers and
 a ≠ 0.
Quadratic equations of this form are said to be in standard form.
Examples: 1: Write each of these quadratic equations in standard form and identify the values of a, b, and

a. 2 x2 −4 x−1=0 This equation is already in standard form

a=2 , b=−4 , c=−1
 b. x 2+ 2 x=3 Add -3 to both sides so that the right-hand side will be
x 2+ 2 x−3=3−3 equal to 0.
x 2+ 2 x−3=0
a=1 ,b=2 , c=−3

c. 6 x 2=9 This time, x is missing. Hence, b = 0.

6 x 2=9 Add -9 to both sides so that the right hand side will be
6 x 2−9=9−9 equal to 0.
6 x 2−9=0
a=6 , b=0 , c=−9

d. −8 x 2+ x=6 Add -6 to both sides

−8 x 2+ x−6=6−6
−8 x 2+ x−6=0 Multiply both sides by -1
−1 (−8 x 2+ x −6 ) =−1(0)
8 x 2+ x−6=0 This equation is already in standard form.
a=8 , b=1, c=−6

Solving Quadratic Equations by Extracting Square Roots

A method that can be used to solve quadratic equations is extracting

square roots. The quadratic equation x 2=c , where c is a real number, has two
solutions x=√ c and x=− √ c or simply x=± √ c .

The following properties are useful in simplifying expressions containing square roots.

a √a
√ ab=√ a × √ b √ =
b √b

25 √ 25 5
Example : √ 12=√ 4 × √ 3=2 √ 3 Example :
√ 4
=
4
=
2

Definition 1.2
Imaginary
Number
The imaginary number i is defined as the square root of −1.
i=√ −1∨i 2=−1 Example: √ −9=√ 9× √−1=3 i
Example 2: Solve each equation by extracting square roots.
Solutions: a. x 2+ 4=0 Add −4 on each side of the equation
x 2+ 4−4=−4
x 2=−4 Extract square roots
√ x 2=√−4
x=± √−4 Apply the property of square roots
x=± √ 4 × √−1 Simplify
x=± 2i

The roots of x 2+ 4=0 are +2 i and −2 i.

b. 2 x2 −11=25 Add 11 on each side of the equation

2 x2 −11+11=25+11
2 x2 =36 Divide each side by 2
2
x =18 Extract square roots
√ x 2=± √18
x=± √ 9 × √ 2 Apply the property of square roots
x=± 3 √ 2

The roots of 2 x2 −11=25 are 3 √ 2 and −3 √ 2.

c. x 2−64=0 Add 64 to each side of the equation

x 2−64+ 64=64
x 2=64 Extract square roots
√ x 2=± √64
x=± 8

The roots of x 2−64=0 are 8 and −8.

d. x 2−50=0 Add 50 to each side of the equation

x 2−50+50=50
x 2=50 Extract square roots
2
√ x =± √50
x=± √ 25× √ 2 Apply the property of square roots
x=± 5 √ 2

The roots of x 2−50=0 are 5 √ 2 and −5 √ 2.

Solving Quadratic Equations by Factoring
A quadratic equation can be solved by factoring. To solve a quadratic
equation using this method, write in the standard form ax 2 +bx +c=0 . Then factor
ax 2 +bx +c=0 and apply the Zero Product Principle.

Steps in Solving Quadratic Equations by Factoring

Transform the given equation to standard form ax 2 +bx +c=0 (if not arranged).
Factor the equation completely.
Equate the factors to zero.
Check your answers.

Example 3: Solve each quadratic equation by factoring.

Solutions: a. x 2+ 3 x =4 Write the equation in standard form

2
x + 3 x −4=0 Factor the left-hand side of the equation
( x +4 )( x−1 )=0
x +4=0 and x−1=0 Equate the factors to zero, then solve each equation
x=−4 and x=1

The roots of x 2+ 3 x −4=0 are −4 and 1.

Check your answers: If x=−4: If x=1:
x 2+ 3 x =4 x 2+ 3 x =4
2 2
(−4) + 3(−4)=4 (1) +3 (1)=4
16−12=4 1+3=4
4=4 (True) 4=4 (True)

b. x 2−8 x +16=0 Factor the left-hand side of the equation

( x−4 ) ( x−4 )=0
x−4=0 and x−4=0 Equate the factors to zero, then solve each equation
x=4 and x=4

The roots of x 2−8 x +16=0 is 4.

Check your answers: If x=4:
x 2−8 x +16=0
( 4)2−8 ( 4 ) +16=0
16−32+ 16=0
−32+32=0
0=0 (True)
 c. x 2=2 x+15 Write the equation in standard form
x 2−2 x−15=0 Factor the left-hand side of the equation
( x−5 ) ( x+ 3 )=0
x−5=0 and x +3=0 Equate the factors to zero, then solve each equation
x=5 and x=−3

The roots of x 2=2 x+15 are 5 and −3.

Check your answers: If x=5: If x=−3:

x 2=2 x+15 x 2=2 x+15
2 2
(5) =2 ( 5 ) +15 (−3) =2 (−3 ) +15
25=10+15 9=−6+15
25=25 (True) 9=9 (True)

Solving Quadratic Equations by Completing the Square

To complete the square of the expression x 2+ bx , add the square of half
b 2
2
the coefficient of x to make x + bx+
2
.()
Example: Complete the square
−8
a. x 2−8 x Half of coefficient x: =−4
2
Square of half of coefficient x: (−4)2=16
Complete the square: x 2−8 x +16
The trinomial x 2−8 x +16 can be written as ( x−4)2 .
3 3
b. x 2+ 3 x Half of coefficient x: =
2 2
3 2 9
Square of half of coefficient x: ( ) =
2 4
2 9
Complete the square: x + 3 x +
4
2 9 3 2
The trinomial x + 3 x + can be written as ( x + ) .
4 2
Steps in Solving Quadratic Equations by Completing the Square
Transpose the constant to the right side of the equation.
If the coefficient of x 2 ≠ 1, divide both members by that coefficient.
Find the 3rd term that make the left side a perfect square trinomial.
Add this term to both sides of the equation.
Express the left side as the square of a binomial and simplify the right side.
Extract the square roots of both sides.
Solve for x. Simplify results if possible. Identify the two roots clearly.

Example 4: Solve each quadratic equation by completing the square.

a. x 2+ 10 x−11=0
b. 2 x2 +8 x−24=0

Solutions: a. x 2+ 10 x−11=0 Transpose the constant to the right side of the equation.

2 10 2 2
x + 10 x=11 Add the square of half of coefficient x: ( ) =5 =25 to
2
x 2+ 10 x +25=11+25 both sides of the equation.
( x +5)2=36
√(x +5)2=√ 36 Extract the square roots of both sides.
x +5=± 6 Simplify
x +5=6 x +5=−6
x=6−5=1 x=−6−5=−11

The roots of x 2+ 10 x−11=0 are 1 and −11.

d. 2 x2 +8 x−24=0 Divide both sides of the equation by 2.

2
x + 4 x−12=0 Transpose the constant to the right side of the equation.
2
2 4 2
x + 4 x=12 Add the square of half of coefficient x: ( ) =2 =4 to
2
x 2+ 4 x +4=12+ 4 both sides of the equation.
( x +2)2 =16
√(x +2)2=√ 16 Extract the square roots of both sides.
 x +2=± 4 Simplify
x +2=4 x +2=−4
x=4−2=2 x=−4−2=−6

The roots of 2 x2 +8 x−24=0 are 2 and −6.

Solving Quadratic Equations using the Quadratic Formula

The quadratic formula is similar to a smartphone in one aspect-- its
versatility. While factoring and extracting square roots are appropriate to use for
solving certain quadratic equations, quadratic formula can be used to find the
solution to any quadratic equation.

The Quadratic Formula is a general formula for solving the quadratic

equation ax 2 +bx +c=0 , where a ≠ 0.

Quadratic formula:

−b ± √ b2−4 ac
If ax 2 +bx +c=0 , where a ≠ 0, then x=
2a

Example 5: Solve each quadratic equation using the quadratic formula.

a. x 2+ 2 x−3=0
b. x 2+ 11 x +28=0

Solutions: a. x 2+ 2 x−3=0 Determine values of a, b, and c.

a=1 ,b=2 , c=−3 Substitute the values of a, b, and c to the formula
−b ± √ b2−4 ac −2 ± √ 22−4(1)(−3)
x= x=
2a 2(1)
−2 ± √ 4−4(−3)
x=
2
−2 ± √ 4 +12
x=
2
−2 ± √ 16
x=
2
−2 ± 4
x=
2
 −2+4 −2 4 2 −2−4 −6
x= = + = =1 x= = =−3
2 2 2 2 2 2
The roots of x 2+ 2 x−3=0 are 1 and −3.

b. x 2+ 11 x +28=0 Determine values of a, b, and c.

a=1 ,b=11, c=28 Substitute the values of a, b, and c to the formula
−b ± √ b2−4 ac −11± √ 11 2−4 (1)(28)
x= x=
2a 2(1)
−11 ± √121−112
x=
2
−11 ± √ 9
x=
2
−11 ±3
x=
2
−11+3 −8 −11−3 −14
x= = =−4 x= = =−7
2 2 2 2
The roots of x 2+ 11 x +28=0 are −4 and −7.