Question

Solve the following inequality:
$\sqrt{(x+3)(x-8)}>x+2.$

Answer 1

$(x+3)(x-8)\ge 0$

$xϵ(-\infty ,-3]$ $\cup [8,\infty )$

$\sqrt{(x+3)(x-8)}>x+2$

Squarring both sides

$x^2-5x-24>x^2+4x+4$

$9x+28<0$

$xϵ(-\infty ,\frac{-28}{9})$