SOLVING QUADRATIC EQUATIONS BY FACTORING WITH 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)/ 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

2x - 3  =  0

2x  =  3

x  =  3/2

2x + 5  =  0

2x  =  -5

x  =  -5/2

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

Example 2 :

Solve the following equation by factoring :

a2b2x2 – (a+ b2)x + 1  =  0

Solution :

a2b2x2 – (a+ 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

b2x – 1  =  0

b2x  =  1

x  =  1/b2

a2x – 1  =  0

a2x  =  1

x  =  1/a2

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

2y + 3  =  0

2y  =  -3

y  =  -3/2

y - 4  =  0

y  =  4

Substitute (x + 1) for y. 

x + 1  =  -3/2

x  =  (-3/2) - 1

x  =  -5/2

x + 1  =  4

x  =  4 - 1

x  =  3

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 

3y + 4  = 0

3y  =  4

y  =  4/3

y - 3  = 0

y  =  3

Substitute (x - 4) for y. 

x - 4  =  4/3

x  =  (4/3) + 4

x  =  16/3

x - 4  =  3

x  =  3 + 4

x  =  7

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

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