Question

Solve the following inequality:
$\frac{x^2-7|x|+10}{x^2-6x+9}<0$

Answer 1

$x\ge 0$

$\frac{x^2-7x+10}{x^2-6x+9}<0$

$\frac{(x-5)(x-2)}{(x-3{)}^2}<0$

$x\in (0,2)\cup (5,\infty )$

And

$x<0$

$\frac{x^2+7x+10}{x^2-6x+9}<0$

$\frac{(x+5)(x+2)}{(x-3{)}^2}<0$

$x\in (-5,-2)$

$\therefore x\in (-5,-2)\cup (0,2)\cup (5,\infty )$