Question

Solve the following inequality:
$|x^2-3x|+x-2<0$

Answer 1

$|x^2-3x|+x-2<0$

$|x^2-3x|<2-x$

$x^2-3x>-(2-x)$ & $x^2-3x<2-x$

$x^2-5x+2>0$ & $x^2-2x-2<0$

$(x-\frac{5+\sqrt{17}}{2})(x-\frac{5-\sqrt{17}}{2})>0$

$xϵ(-\infty ,\frac{-5-\sqrt{17}}{2})\cup (\frac{5+\sqrt{17}}{2},\infty )$

$x^2-2x-2<0$

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

$(x-1+\sqrt{3})(x-1-\sqrt{3})<0$

$xϵ(1-\sqrt{3},1+\sqrt{3})$

So, $xϵ\varphi$