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
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 – (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
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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