Question

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

Answer 1

\(= (x-1)(x-4) = 18

= x^2-4x-x+4 =18

= x^2-5x-14=0

= x^2-7x+2x-14=0

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

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

Either $x-7=0$

Therefore$x=7$

Or,$x+2=0$

Therefore $x=-2$

The roots of the above mentioned quadratic equation are 7 and -2 respectively