Solving Quadratic Equations

2. Solving Quadratic Equations
by Extracting the Square Root,
Factoring, Completing the
Square, and using the
Quadratic Formula
MARK JOVEN A. ALAM-ALAM, LPT

3. Quadratic Equation
 can have two real solutions, one
real solution, or no real solution

4. Extracting the Square Root
 can be used to solve a quadratic equation written in the form 𝒙𝟐 = c
based on the square root property, can be done by extracting the square
roots of both sides of the equation
𝒙𝟐
= c
 has exactly two solutions :
x = ± 𝑐 or x = 𝑐 and x = - 𝑐
 If c > 0, then 𝑥2 = c has two real solutions, if c=0, then the only solution
to the equation is 0

5. Example 1: Find the roots of each quadratic
equation by extracting the square root.
a. 𝑥2 = 64
b. 𝑥2 - 100 = 0
c. -4𝑥2
+ 100 = 0

6. Solution:
a. 𝑥2
= 64
𝑥2 = ± 64
𝑥 = ± 8
x = 8 or x = -8
Checking:
If x = 8
𝑥2
= 64
(8)2
= 64
64 = 64
If x = -8
𝑥2
= 64
(−8)2
= 64
64 = 64

7. Solution:
b. 𝑥2
- 100 = 0
x2
= 100
𝑥2 = ± 100
𝑥 = ± 10
x = 10 or x = -10
Checking:
If x = 10
𝑥2
- 100 = 0
(10)2
- 100 = 0
100 -100 = 0
0 = 0
If x = -10
𝑥2
- 100 = 0
(−10)2
- 100 = 0
100 -100 = 0
0 = 0

8. Solution:
c. -4𝑥2
+ 100 = 0
−4x2
= -100
−4x2
−4
=
−100
−4
x2
= 25
𝑥2 = ± 25
𝑥 = ± 5
x = 5 or x = -5
Checking:
If x = 5
-4𝑥2
+100 = 0
-4(5)2
+ 100 = 0
-4(25)+100 = 0
-100+100 = 0
0 = 0
If x = -5
-4𝑥2
+100 = 0
-4(−5)2
+ 100 = 0
-4(25)+100 = 0
-100+100 = 0
0 = 0

9. Take Note!!!
 In cases that c<0 in the quadratic
equation 𝑥2 = c, there is no real number
solution to the quadratic equation.
 In this case, the quadratic equation
has imaginary roots
x = ± i 𝑐 where i = −1

10. Factoring
 used to find the solutions of a quadratic equation in the
form a𝑥2
+ bx + c = 0, where a is not equal to 0
 based on the zero factor property, which states that if a
and b are algebraic expressions, then the equation ab=0 is
equivalent to the compound statement a=0 or b=0
Quadratic equation in the form a𝒙𝟐
+ bx + c = 0
 factorable if discriminant (𝑏2
- 4ac) is a perfect square

11. Example 2: Solve the following quadratic
equations by factoring.
a. 𝑥2
+ 7x = 0
b. 4𝑥2
= 6x
c. -5𝑥2
+ 10x = 20x

12. Solution:
a. 𝑥2
+ 7x = 0
x(x + 7) = 0
x = 0 x + 7= 0
x = -7

13. Solution:
b. 4𝑥2
= 6x
4𝑥2
- 6x = 0
2x(2x – 3) = 0
2x = 0 2x - 3= 0
x=0 2x = 3
x=
3
2

14. Solution:
c. -5𝑥2
+ 10x = 20x
-5𝑥2
+ 10x -20x = 0
-5𝑥2
- 10x = 0
-5x(x + 2) = 0
-5x = 0 x + 2= 0
x=0 x = -2

15. Completing the Square
 not all quadratic equations contain a perfect
square trinomial in any of the sides
one must make the left side of the quadratic
equation a perfect square trinomial in the form
𝑥2
+ 2bx + 𝑏2
= (𝑥 + 𝑏)2

16. Steps in Solving Quadratic Equations by
Completing the Square:
1. Rewrite the equation in the standard form of the quadratic equation, 𝑎𝑥2
+ bx = c.
2. Make sure that a = 1. If it is not, then divide each term of the equation by the value
of a.
3. Take half of the coefficient b and square the quotient. Add the result to both sides of
the equation. The number added to both sides of the equation serves as the constant
term of the trinomial on the left side. This makes the left side of the equation a perfect
trinomial.
4. Factor the perfect square trinomial on the left side of the equation and express it as a
square binomial.
5. Extract the square root of both sides to determine the solutions of the given
quadratic equation. Remember to consider the positive and negative results of the right
side of the equation.

17. Example 3: Solve the following quadratic
equations by completing the square.
a. 𝑥2
- 8x + 15 = 0
b. 2𝑥2
+ 4x = 16

18. Solution:
a. 𝑥2
- 8x + 15 = 0
𝑥2 - 8x = -15
𝑥2
- 8x + 16= - 15 + 16
𝑥2 - 8x + 16 = 1
(𝑥 − 4)2 = 1
(𝑥 − 4)2 = ± 1
x-4 = ± 1
x - 4 = 1 x -4 = -1
x= 4 + 1 x = 4-1
x = 5 x = 3

19. Solution:
b. 2𝑥2
+ 4x = 16
2𝑥
2
2
+
4𝑥
2
=
16
2
𝑥2 + 2x = 8
𝑥2
+ 2x + 1 = 8 + 1
(𝑥 + 1)2 = 9
(𝑥 + 1)2 = ± 9
x + 1 = ± 3
x + 1 = 3 x + 1 = -3
x = 3 - 1 x = -3 - 1
x = 2 x = -4

20. Quadratic Formula
x =
−𝑏 ± 𝑏2−4𝑎𝑐
2𝑎

21. Example 4: Solve the following quadratic
equations by using the quadratic formula.
a. 𝑥2
+ 11x + 24 = 0
b. (2x-1)(x-6) = 1 - 2x

22. Solution:
a. 𝑥2
+ 11x + 24 = 0
a=1 b=11 c=24
x =
−𝑏 ± 𝑏2−4𝑎𝑐
2𝑎
x =
−11 ± 112−4(1)(24)
2(1)
x =
−11 ± 121−96
2
x =
−11 ± 25
2
x =
−11+ 25
2
x =
−11 +5
2
x =
−6
2
x = -3
x =
−11− 25
2
x =
−11 −5
2
x =
−16
2
x = -8

23. Solution:
b. (2x-1)(x-6) = 1 - 2x
2𝑥2 -12x –x + 6 = 1 – 2x
2𝑥2 -12x –x + 2x + 6 - 1= 0
2𝑥2 -11x + 5 = 0
a=2 b=-11 c=5
x =
−𝑏 ± 𝑏2−4𝑎𝑐
2𝑎
x =
−(−11) ± −112−4(2)(5)
2(2)
x =
11 ± 121−40
4
x =
11 ± 81
2
x =
11+ 81
4
x =
11 +9
4
x =
20
4
x = 5
x =
11− 81
4
x =
11 −9
4
x =
2
4
x =
1
2