Question

Solve the following inequality:
$a\sqrt{x+1}<1.$

Answer 1

$a\sqrt{x+1}<1$

if $a>0\sqrt{x+1}<\frac{1}{a}x+1<\frac{1}{a^2};x<\frac{1}{a^2}-1$

if $a<0\sqrt{x+1}>\frac{1}{a}x>\frac{1}{a^2}-1$.