Question

Solve the following inequality:
$4-x<\sqrt{x^2-2x}.$

Answer 1

$x^2-2x\ge 0$

$x(x-2)\ge 0$

$xϵ(-\infty ,0]\cup [2,\infty )$

$4-x<\sqrt{x^2-2x}$

Squarring both sides

$16+x^2-8x<x^2-2x$

$6x-16>0$

$3x-8>0$

$xϵ(\frac{8}{3},\infty )$