Question

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

Answer 1

$4-x^2\ge 0$

$x^2-4\le 0$

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

$xϵ[-2,2]$

$x+2>2\sqrt{4-x^2}$

Squarring both sides

$x^2+4x+4>16-4x^2$

$5x^2+4x-12>0$

$5x^2+10x-6x-12>0$

$(5x-6)(x+2)>0$

$xϵ(-\infty ,-2)\cup (\frac{6}{5},\infty )$

$xϵ(\frac{6}{5},2]$