Question

Solve the following inequality:
$\sqrt{2x^2-3x-5}<x-1.$

Answer 1

$2x^2-3x-5\ge 0$

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

$(2x-5)(x+1)\ge 0$

$xϵ(-\infty ,-1]\cup [\frac{5}{2},\infty )$

$\sqrt{2x^2-3x-5}<x-1$

Squarring both sides

$2x^2-3x-5<x^2-2x+1$

$x^2-x-6<0$

$(x-3)(x+2)<0$

$xϵ(-2,3)$

$xϵ(-2,-1]\cup [\frac{5}{2},3)$