Lesson 1 - Math 9-W1Q1_Solving Quadratic Equations by Factoring and by completing the square

1. LESSON 1
SOLVING QUADRATIC
EQUATIONS BY
FACTORING AND BY
COMPLETING THE
SQUARE
GRADE 9

2. Quadratic Equation
• In terms of x is a second-degree polynomial equation written in standard
form as:
𝑎𝑥2
+ bx + c = 0
• The quadratic term of a quadratic equation cannot be equal to zero.
To solve quadratic equations by factoring, you must follow these steps:
1. Transform the quadratic equation in standard form.
2. Factor the polynomial side of the equation.
3. Use the Zero-Product property to equate each factor to zero.
4. Solve the resulting equations.
5. Check if the obtained roots are accurate by substituting them to the original equations.

3. Example 1. Solve the quadratic equation 𝒙𝟐
= 6x + 27
Solution: 𝒙𝟐 = 6x + 27
𝒙𝟐- 6x – 27 = 0 - Transform the given to standard form
𝒙 − 𝟗 𝒙 + 𝟑 = 𝟎 - Factor the polynomial side of the equation
x - 9 = 0 or x +3 = 0 - Use the Zero-Product Property
x = 9 x = -3
Checking:
If x = 9
𝟗𝟐
= 6(9) + 27
𝟗𝟐
= 54 + 27
81= 81
If x = -3
−𝟑𝟐
= 6(-3) + 27
−𝟑𝟐
= -18+ 27
𝟗 = 9

4. Example 2. Solve the quadratic equation 𝟓(𝒚𝟐
+ 7) = -3 (5y-8) + 2y + 17
S𝐨𝐥𝐮𝐭𝐢𝐨𝐧: 𝟓(𝒚𝟐
+ 7) = -3 (5y - 8) + 2y + 17
𝟓(𝒚𝟐 + 7) = -3 (5y - 8) + 2y + 17
𝟓𝒚𝟐 + 35 = -15y + 24 + 2y + 17
𝟓𝒚𝟐 + 35 = -15y + 2y + 24 + 17
𝟓𝒚𝟐 + 35 = -13y + 41
𝟓𝒚𝟐
+ 13y+ 35-41 = 0
𝟓𝒚𝟐
+ 13y - 6= 0
𝒚 + 𝟑 = 0
𝒚 + 𝟑= 0
𝒚 = -3
𝟓𝒚 − 𝟐= 0
𝟓𝒚 = 2
𝟓𝒚 = 2
5 5
𝒚 =
𝟐
𝟓

5. Checking using the standard form of the equations
If 𝒚 = -3
𝟓𝒚𝟐 + 13y - 6= 0
𝟓(−𝟑)𝟐
+ 13(-3) - 6= 0
𝟓(𝟗) - 39 - 6= 0
𝟒𝟓 - 45= 0
𝟎= 0
If 𝒚 =
𝟐
𝟓
5(
𝟐
𝟓
)𝟐
+ 13 (
𝟐
𝟓
) – 6 = 0
5(
𝟒
𝟐𝟓
) +
𝟐𝟔
𝟓
– 6 = 0
𝟐𝟎
𝟐𝟓
+
𝟐𝟔
𝟓
– 6 = 0
𝟐𝟎
𝟐𝟓
+
𝟐𝟔
𝟓
=
100 650
125
𝟕𝟓𝟎
𝟏𝟐𝟓
-6= 0
0 = 0

6. Example 3: Solve the quadratic equation
𝟏
𝟏𝟎
𝒂𝟐 -
𝟕
𝟏𝟎
a -12 = 0
Solution:
𝟏
𝟏𝟎
𝒂𝟐
-
𝟕
𝟏𝟎
a -12 = 0
10 (
𝟏
𝟏𝟎
𝒂𝟐
-
𝟕
𝟏𝟎
a -12) =10 ( 0 ) > Transform the numerical coefficients to integral
coefficients
𝒂𝟐 - 𝟕a - 120 = 0
𝒂 − 𝟏𝟓 (𝒂 + 𝟖) = 0 > Use the Zero-Product Property
a – 15 = 0
a = 15
a + 8 = 0
a = -8
Checking: 𝟏𝟓𝟐
- 𝟕(15) - 120 = 0
𝟏𝟓𝟐
- 𝟕(15) - 120 = 0
𝟐𝟐𝟓 − 𝟏𝟎𝟓 − 𝟏𝟐𝟎 = 𝟎
𝟎 = 𝟎
−𝟖𝟐
- 𝟕(-8) - 120 = 0
64 +56 - 120 = 0
120 - 120 = 0
0 = 0

7. Example 4: Solve the quadratic equation 𝟑𝒙𝟐
+ 𝟖𝒙 - 11 = 0
Solution: 𝟑𝒙𝟐
+ 𝟖𝒙 - 11 = 0
𝟑𝒙𝟐 + 𝟖𝒙 - 11 = 0
𝟑𝒙 + 𝟏𝟏 𝒙 − 𝟏 = 𝟎
𝟑𝒙𝟐 - 3𝒙 + 11x - 11 = 0
𝟑𝒙𝟐
+ 8x - 11 = 0
3x + 11= −𝟏𝟏
3x = −𝟏𝟏
3 = 𝟑 x=
−𝟏𝟏
𝟑
x - 1= 𝟏
x = 𝟏

8. Practice Exercises:
Solve the quadratic equation by factoring:
1. 𝟒𝒙𝟐
- 4x + 1= 0
2. 𝒙𝟐 + 8x = -7
3. 𝒖𝟐
- 15u -34 =0

9. Solving Quadratic Equations Using The Square Root Property
Square Root Property
If 𝑥2 = n , then x = + 𝑛
a. 𝑥2 = 25 𝑥2 = 25
𝑥2 = 25
x = 25
x = + 5
𝑥2
= 121
b. Solve the quadratic equation 𝑥2
-121= 0
𝑥2 = 121
𝑥 = ± 11
Checking:
If x = 11
112
-121 = 0
121-121 = 0
0 = 0
Thus, the solution set of
the quadratic equation
is (-5,5)
Thus, the solution set of
the quadratic equation
is (-11,11)

10. Practice Exercises
1. 𝑡2
− 4 = 12
2. 𝑥2
− 400 = 0
3. 7𝑎2
− 300 = 43

11. Example 1: Solve the quadratic equation 𝒙𝟐
+ 6x - 5= 0
Solution:
𝒙𝟐 + 6x - 5= 0
𝒙𝟐
+ 6x = 5 transform the given to 𝑎𝑥2
+ bx = c
𝒙𝟐 + 6x + 9 = 5 + 9 - Step 2 will not be necessary since a = 1 . Next, add
the square of one-half of the middle term to both sides
of the equation.
Half of 6 is 3 and its square is 9
( 𝒙 + 𝟑 )𝟐
= 14 Factor the left side resulting to a square of binomial
( 𝒙 + 𝟑 )𝟐= 𝟏𝟒 Solve the resulting equation by extracting the square root
Subtract 5 to both sides
Solving Quadratic Equations by Completing the Square

12. x + 3= 𝟏𝟒
x = -3 ± 𝟏𝟒
x = -3 + 𝟏𝟒 or x = -3 - 𝟏𝟒
Thus, the solution set of the quadratic equation
𝒙𝟐 + 6x - 5= 0 is (-3 + 𝟏𝟒 , -3 - 𝟏𝟒)

13. Example 2: Solve the quadratic equation 𝒂𝟐
- 3a + 9= 0
Solution: 𝒂𝟐
- 3a + 9= 0
𝒂𝟐
- 3a = -9
𝒂𝟐 - 3a +
𝟗
𝟒
= -9 +
𝟗
𝟒
(a −
𝟑
𝟐
)
𝟐
= -
𝟐𝟕
𝟒
(a −
𝟑
𝟐
)
𝟐
= −
𝟐𝟕
𝟒
a −
𝟑
𝟐
= −
𝟐𝟕
𝟒
a −
𝟑
𝟐
= ±
𝟑 𝟑
𝟐
a =
𝟑
𝟐
±
𝟑 𝟑
𝟐
a =
𝟑 ±𝟑 𝟑
𝟐
Thus, the roots of the quadratic equation
𝒂𝟐
- 3a + 9= 0 are {
𝟑 + 𝟑 𝟑
𝟐
,
𝟑 −𝟑 𝟑
𝟐
}

14. Practice Exercises:
Solve the following quadratic equation by completing the square:
1. 𝒂𝟐
- 6a -16 = 0
2. 𝒄𝟐
+ 2c = 48
3. 𝒆𝟐
= -7e + 30