Question

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

Answer 1

$x^2+x-2\ge 0$

$(x+2)(x-1)\ge 0$

$xϵ(-\infty ,-2)\cup [1,\infty )$

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

Squarring both sides

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

$x-2>0$

$xϵ(2,\infty )$