Question

Solve the following quadratic equations by factorization:

${\displaystyle \frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}}$

Answer 1

We have been given,

${\displaystyle \frac{1}{x-1}(x-2)+\frac{1}{x-2}(x-3)+\frac{1}{x-3}(x-4)=\frac{1}{6}}$

${\displaystyle \frac{(x-3)(x-4)+(x-1)(x-4)+(x-1)(x-2)}{(x-1)(x-4)(x-2)(x-3)}=\frac{1}{6}}$

${\displaystyle \frac{3(x^2-5x+6)}{(x^2-5x+4)(x^2-5x+6)}=\frac{1}{6}}$

18 = x² - 5x + 4

x² - 5x - 14 = 0

x² - 7x + 2x - 14 = 0

x(x - 7) + 2(x - 7) = 0

(x + 2)(x - 7) = 0

Therefore,

x + 2 = 0

x = -2

or,

x - 7 = 0

x = 7

Hence, x = -2 or x = 7.