Question

Solve the following inequality:
$2x+3<\sqrt{x^2+5x+6}.$

Answer 1

$x^2+5x+6\ge 0$

$(x+2)(x+3)\le 0$

$xϵ(-\infty ,-3]\cup [-2,\infty )$

$2x+3<\sqrt{x^2+5x+6}$

Squarring both sides

$4x^2+12x+9<x^2+5x+6$

$3x^2+7x+3<0$

$(x+\frac{7+\sqrt{13}6}{6})(x+\frac{7-\sqrt{13}}{6})<0$

$xϵ(\frac{-7-\sqrt{13}}{6},\frac{-7+\sqrt{13}}{6})$