Question

Solve the following inequality:
$\sqrt{x^2+2x-3}<1.$

Answer 1

$\sqrt{x^2+2x-3}<1$

$\sqrt{x^2+2x-3}$

$x^2+2x-3\ge 0$

$(x-1)(x+3)\ge 0$

$x\le -3orx\ge 1$

$\sqrt{x^2+2x-3}<1$

$x^2+2x-3<1$

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

$(x-(\sqrt{5}-1))(x-(-1-\sqrt{5}))<0$

$-1-\sqrt{5}<x<\sqrt{5}-1$

$-1-\sqrt{5}<x\le -3or1\le x<\sqrt{5}-1$