Solve the following quadratic equations by factorization method :
$\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6}$

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Solution

$\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6} \Rightarrow \frac{(x - 3) (x - 4) + (x - 1) (x - 4) + (x - 1) (x - 2)}{(x - 1) (x - 2) (x - 3) (x - 4)} = \frac{1}{6} \Rightarrow 6 (x - 7 x + 12 + x^{2} - 5 x + 4 + x^{2} - 3 x + 7) = (x - 1) (x - 2) (x - 3) (x - 4) \Rightarrow 6 (3 x^{2} - 15 x + 18) = (x - 1) (x - 2) (x - 3) (x - 4) \Rightarrow 6 \times 3 (x^{2} - 5 x + 6) = (x - 1) (x - 2) (x - 3) (x - 4) \Rightarrow 18 (x^{2} - 3 x - 2 x + 6) = (x - 1) (x - 2) (x - 3) (x - 4) \Rightarrow (x - 1) (x - 4) = 18 \Rightarrow x^{2} - 5 x + 4 - 18 = 0 \Rightarrow x^{2} - 7 x + 2 x - 14 = 0 \Rightarrow (x - 7) (x + 2) = 0 \Rightarrow x - 7 = 0 o r, x + 2 = 0 x = 7 o r x = - 2$
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