Question

Solve the following inequality:
$x-3<\sqrt{x^2+4x-5}.$

Answer 1

$x^2+4x-5\ge 0$

$(x+5)(x-1)\ge 0$

$xϵ(-\infty ,-5)\cup [1,\infty )$

$(x-3)<\sqrt{x^2+4x-5}$

Squarring both sides

$x^2-6x+9<x^2+4x-5$

$10x-14>0$

$5x-7>0$

$xϵ(\frac{7}{5},\infty )$