Question

Solve the following quadratic equation by factorization, the sum of the roots are :
$\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}$

Answer 1

Given,

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

$6(x-3)(x-4)+6(x-1)(x-4)+6(x-1)(x-2)=(x-1)(x-2)(x-3)(x-4)$

$6\left[\right(x-3\left)\right(x-4)+(x-1\left)\right(x-4)+(x-1\left)\right(x-2\left)\right]=(x-1)(x-2)(x-3)(x-4)$

$6[x^2-7x+12+x^2-5x+4+x^2-3x+2]=(x-1)(x-2)(x-3)(x-4)$

$6(3x^2-15x+18)=x^4-10x^3+35x^2-50x+24$

$18x^2-90x+108=x^4-10x^3+35x^2-50x+24$

solving the above equation, we get,

$x^4-10x^3+17x^2+40x-84=0$

$(x+2)(x-2)(x-3)(x-7)=0$

$\text{as undefined points are}$ $2,3$

$\therefore x=-2,7$
$\text{sum of roots=(-2)+(+7)}=5$