Question

Solve the following inequality:
$x+4<\sqrt{-x^2-8x-12}.$

Answer 1

$-x^2-8x-12\ge 0$

$x^2+8x+12\le 0$

$(x+6)(x+2)\le 0$

$xϵ[-6,-2]$

$x+4<\sqrt{-x^2-8x-16}$

Squarring both sides

$x^2+8x+16<-x^2-8x-16$

$x^2+8x+16<0$

$(x+2)(x+6)<0$

$xϵ(-6,-2)$