Question

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

Answer 1

$x^2+x\ge 0$

$x(x+1)\ge 0$

$xϵ(-\infty ,-1]\cup [0,\infty )$

$\sqrt{x^2+x}>1-2x$

Squarring both sides

$x^2+x>4x^2-4x+1$

$3x^2-5x+1<0$

$xϵ(\frac{5+\sqrt{13}}{6},(\frac{-5+\sqrt{13}}{6})$

$xϵ(\frac{-5-\sqrt{13}}{6},-1]$