SOLVING QUADRATIC EQUATIONS BY FACTORING WITH EXAMPLES

Solving Quadratic Equations by Factoring with Examples :

In this section, we will learn how to solve quadratic equation by factoring method.. 

Solving Quadratic Equations by Factoring Method Examples

Example 1 :

Solve the following equation by factoring :

[x/(x + 1)] + [(x + 1)/x]  =  34/15

Solution :

[x/(x + 1) ] + [(x + 1)/x]  =  34/15

[x² + (x + 1)² / x(x + 1)]  =  34/15

(x² + x² + 2x + 1) / (x² + x)  =  34/15

15(2x² + 2x + 1)  =  34(x² + x)

30x² + 30x + 15  =  34x² + 34 x

34x² - 30x² + 34x – 30x – 15  =  0

4x² + 4x – 15  =  0

4x² + 10x - 6x – 15  =  0

2x(2x + 5) – 3(2x + 5)  =  0

(2x – 3)(2x + 5)  =  0

2x - 3  =  0  or 2x + 5  =  0

So,the solution is {-5/2, 3/2}.

Example 2 :

Solve the following equation by factoring :

a²b²x² – (a² + b²)x + 1  =  0

Solution :

a²b²x² – (a² + b²)x + 1  =  0

a²b²x² – a²x - b²x + 1  =  0

a²x(b²x – 1) -1(b²x – 1)  =  0

(b²x – 1)(a²x – 1)  =  0

b²x – 1  =  0  or  a²x – 1  =  0

So, the solution is { 1/a², 1/b² }. 

Example 3 :

Solve the following equation by factoring :

2(x + 1)² – 5(x + 1) = 12

Solution :

2(x + 1)² – 5(x + 1) = 12

Let y  =  x + 1.

Then,

2y² – 5 y = 12

2y² – 5y – 12 = 0

2y² – 8y + 3y – 12 = 0

2y (y – 4) + 3 (y – 4) = 0

(2y + 3)(y – 4) = 0

2y + 3  =  0  or  y - 4  =  0

Substitute (x + 1) for y. 

So, the solution is { -5/2, 3 }.

Example 4 :

Solve the following equation by factoring :

3(x – 4)² – 5(x – 4)  =  12

Solution :

3(x – 4)² – 5(x – 4)  =  12

Let y = x – 4.

Then, 

3y² – 5y  =  12

3y² – 5y – 12  =  0

3y² – 9y  +  4y – 12  =  0

3y(y – 3) +  4(y – 3) = 0

(3y + 4)(y – 3) = 0

3y + 4  =  0  or y - 3  =  0 

Substitute (x - 4) for y. 

So, the solution is { 16/3, 7 }.

After having gone through the stuff given above, we hope that the students would have understood how to solve quadratic equations using factoring method.

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