Question

Solve the following inequality:
$\sqrt{(x+4)(x+3)}>6-x.$

Answer 1

$(x+4)(x+3)\ge 0$

$xϵ(-\infty ,-3]\cup [-4,\infty )$

$\sqrt{(x+4)(x+3)}>6-x$

Squarring both sides

$x^2+7x+12>x^2-12x+36$

$19x-24>0$

$xϵ(\frac{24}{19},\infty )$