Question

Solve the following inequality:
$\sqrt{5-x^2}>x-1.$

Answer 1

$5-x^2\ge 0$

$x^2-5\le 0$

$(x-\sqrt{5})(x+\sqrt{5})\le 0$

$xϵ[-\sqrt{5},\sqrt{5}]$

$\sqrt{5-x^2}>x-1$

Squarring both sides

$5-x^2>x^2-2x+1$

$2x^2-2x-4<0$

$x^2-x-2<0$

$(x-2)(x+1)<0$

$xϵ(-1,2)$