SOLVING QUADRATIC EQUATIONS USING FACTORING

Generally all quadratic equation will be in the form

ax² + bx + c = 0

Using factoring method to solve a quadratic equation,  we have to check whether the coefficient of x² is 1 or not equal to 1. 

Solve and check each of the equations.

Problem 1 :

x² - 7x + 10 = 0

Solution :

x² - 7x + 10 = 0

x² - 5x – 2x + 10 = 0

x(x - 5) - 2(x - 5) = 0

(x - 2) (x - 5) = 0

x = 2 or x = 5

If those values will satisfy the equation, we can say these are solutions.

Solve the following quadratic equation :

Problem 2 :

x² - 5x - 6 = 0

Solution :

x² - 5x - 6 = 0

x² + x – 6x – 6 = 0

x(x + 1) – 6(x + 1) = 0

(x + 1) (x - 6) = 0

x = -1 or x = 6

Problem 3 :

x² + 6x + 5 = 0

Solution :

x² + 6x + 5 = 0

x² + x + 5x + 5 = 0

x(x  + 1) + 5 (x + 1) = 0

(x + 1) (x + 5) = 0

x = -1 or x = -5

Problem 4 :

x² + 10x – 24 = 0

Solution :

x² + 10x - 24 = 0

x² - 2x + 12x - 24 = 0

x(x - 2) + 12(x - 2) = 0

(x - 2) (x + 12) = 0

x = 2 or x = -12

Problem 5 :

2x² - x = 12 + x

Solution :

2x² - x = 12 + x

2x² - x - x - 12 = 0

2x² - 2x - 12 = 0

2(x² - x - 6) = 0

2(x² + 2x - 3x - 6) = 0

2(x + 2) (x - 3) = 0

(x + 2) (x - 3) = 0

x = -2 or x = 3

Solving Quadratic Equations not in Standard Form

Problem 6 :

x² - 9x = 10

Solution :

x² - 9x = 10

x² - 9x - 10 = 0

x² + x - 10x - 10 = 0

x(x + 1) - 10(x + 1) = 0

(x + 1) (x - 10) = 0

x = -1 or x = 10

Problem 7 :

4 - x(x - 3) = 0

Solution :

4 - x(x - 3) = 0

4 - x² + 3x = 0

-x² + 3x + 4 = 0

-x² - x + 4x + 4 = 0

-x(x + 1) + 4(x + 1) = 0

(x + 1) (-x + 4) = 0

x = -1 or x = 4

Problem 8 :

x(x + 7) - 2 = 28

Solution :

x(x + 7) - 2 = 28

x² + 7x - 2 - 28 = 0

x² + 7x - 30 = 0

x² - 3x + 10x - 30 = 0

x(x - 3) + 10(x - 3) = 0

(x - 3) (x + 10) = 0

x = 3 or x = -10

Problem 10 :

3x² - 5x = 36 - 2x

Solution :

3x² - 5x = 36 - 2x

3x² - 5x + 2x - 36 = 0

3x² - 3x - 36 = 0

3(x² - x - 12) = 0

3(x + 3) (x - 4) = 0

(x + 3) (x - 4) = 0

x = -3 or x = 4

Problem 11 :

7 = x(8 - x)

Solution :

7 = x(8 - x)

7 = 8x - x²

x² - 8x + 7 = 0

(x - 1) (x - 7) = 0

x = 1 or x = 7

Problem 12 :

9 = x(6 - x)

Solution :

9 = x(6 - x)

9 = 6x - x²

-x² + 6x - 9 = 0

x² - 6x + 9 = 0

(x - 3) (x - 3) = 0

x = 3

Problem 13 :

2x(x + 1) = 12

Solution :

2x(x + 1) = 12

2x² + 2x - 12 = 0

2(x² + x - 6) = 0

2(x + 3) (x - 2) = 0

(x + 3) (x - 2) = 0

x = -3 or x = 2

Problem 14 :

x(x - 2) + 2 = 1

Solution :

x(x - 2) + 2 = 1

x² - 2x + 1 = 0

(x - 1) (x - 1) = 0

x = 1

Problem 15 :

3x(x - 10) + 80 = 5

Solution :

3x(x - 10) + 80 = 5

3x² - 30x + 75 = 0

3(x² - 10x + 25) = 0

(x² - 10x + 25) = 0

(x - 5) (x - 5) = 0

x = 5