## About "Solving quadratic equations by factoring examples"

Solving quadratic equations by factoring examples :

Here we are going to see some example problems on solving quadratic equations using the method factoring.

## Solving quadratic equations by factoring in the form ax²+ bx + c

Whenever we have a quadratic equation in the form ax² + bx + c and we need to factor this, first we have to check whether the coefficient of x² is 1 or not.

## Solving quadratic equations in which the value of a is 1

• If it is 1, then we have to take the constant term and split it into two factors.
• In which the product of two factors must be equal to the constant term and the simplified value of those factors equal to the middle term, that is coefficient of x.
• Write each factors is in the form (x + a) (x + b) according to the sign.
• Set each factors equal to zero to get the value of x.

## Solving quadratic equations in which the value of a is not 1

• Multiply the coefficient of x² and constant term.
• Split this product into two factors such that their sum is equal to the coefficient of x .
• Divide each factors by the coefficient of x².
• If it is possible we can simplify, otherwise we have to write the denominator along with x.
• write each factors in the form (x + a) (x + b).
• Set each factors equal to zero to get the value of x.

## Solving quadratic equations by factoring examples

Example 1 :

Solve x² + 9 x + 14  =  0

Solution :

Since the coefficient of x² is 1, split the constant term that into two parts.

14  =  ⋅ 7, 2 + 7  =  9

x² + 9 x + 14  =  0

(x + 2) (x + 7)  =  0

x - 2  =  0     x + 7  =  0

x  =  2, x  =  -7

Hence the solution is {2, -7}

Let us look into the next example problem on "Solving quadratic equations by factoring examples".

Example 2 :

Solve x² - 9 x + 14  =  0

Solution :

Since the coefficient of x² is 1, split the constant term that into two parts.

14  =  -2 ⋅ (-7) , -2 - 7  =  -9

x² - 9 x + 14  =  (x - 2) (x - 7)

(x - 2) (x - 7)  =  0

x - 2  =  0     x - 7  =  0

x  =  2, x  =  7

Hence the solution is {2, 7}

Let us look into the next example problem on "Solving quadratic equations by factoring examples".

Example 3 :

Solve x² + 2 x - 15  =  0

Solution :

Since the coefficient of x² is 1, split the constant term that into two parts.

-15  =  -3 ⋅ 5 , -3 + 5  =  2

x² + 2 x - 15  =  (x - 3) (x + 5)

(x - 3) (x + 5)  =  0

x - 3  =  0     x + 5  =  0

x  =  3, x  =  -5

Hence the solution is {3, -5}

Example 4 :

Solve x² - 2 x - 15  =  0

Solution :

Since the coefficient of x² is 1, split the constant term that into two parts.

-15  =  -5 ⋅ 3 , -5 + 3  =  -2

x² - 2 x - 15  =  (x - 5) (x + 3)

(x - 5) (x + 3)  =  0

x - 5  =  0     x + 3  =  0

x  =  5, x  =  -3

Hence the solution is {-3, 5}

Example 5 :

Solve 2x² + 15 x + 27  =  0

Solution :

Since the coefficient of x² is not 1, multiply the coefficient of x² by the constant term and split it into two parts.

⋅ 27  =  54

54  =  6 ⋅ 9 , 6 + 9  =  15

(x + 3) (2x + 9)  =  0

x + 3  =  0     2x + 9  =  0

x  =  -3, x  =  -9/2

Hence the solution is {-3, -9/2}

Example 6 :

Solve 2x² - 15 x + 27  =  0

Solution :

Since the coefficient of x² is not 1, multiply the coefficient of x² by the constant term and split it into two parts.

⋅ 27  =  54

54  =  -6 ⋅ (-9) , -6 - 9  =  -15

2x² - 6 x - 9x  + 27  =  0

Factor 2x from the first two terms and factor 9 from the third and fourth terms

2x (x - 3) - 9(x - 3)  =  0

(2x - 9) (x - 3) = 0

2x - 9  = 0,  x - 3  =  0

2x = 9

x  =  9/2 and x = 3

Hence the solution is {9/2, 3)

Example 7 :

Solve 2x² + 15 x - 27  =  0

Solution :

Since the coefficient of x² is not 1, multiply the coefficient of x² by the constant term and split it into two parts.

⋅ (-27)  =  -54

-54  =  18 ⋅ (-3) , 18 - 3  =  15

2x² + 18 x - 3 x - 27  =  0

Factor 2x from the first two terms and factor 3 from the third and fourth terms

2x (x - 9) - 3(x - 9)  =  0

(2x - 3) (x - 9) = 0

2x - 3  = 0,  x - 9  =  0

2x = 9

x  =  9/2 and x = 3

Hence the solution is {9/2, 3)

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