Question

Solve the inequality:

$\frac{x^2-x-6}{x^2-6x}\ge 0$

Answer 1

$\frac{x^2-x-6}{x^2-6x}\ge 0$

$\frac{x^2-3x+2x-6}{x^2-6x}\ge 0$

$\frac{x(x-3)+2(x-3)}{x(x-6)}\ge 0$

$\frac{(x+2)(x-3)}{x(x-6)}\ge 0$

$\frac{(x+2)(x-3)x(x-6)}{x^2{(x-6)}^2}\ge 0$

$x(x+2)(x-3)x(x-6)\ge 0.$ ( $\therefore$square of any quantity is > $0$)

$\therefore x\ge 0,x\ge -2,x\le 3,x\le 6$