Question

Solve the following inequality:
$1-x<\sqrt{x^2-2x}.$

Answer 1

$x^2-2x\ge 0$

$x(x-2)\ge 0$

$xϵ(-\infty ,0]\cup [2,\infty )$

$1-x<\sqrt{x^2-2x}$

Squarring both sides

$x^2-2x+1<x^2-2x$

$1<0$

$xϵ\varphi$