Question

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

Answer 1

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

$xϵR$(Discriminant$<0$)

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

Squarring both sides

$x^2+3x+3<4x^2+4x+1$

$3x^2+x-2>0$

$3x^2+3x-2x-2>0$

$3x(x+1)-2(x+1)>0$

$(3x-2)(x+1)>0$

$xϵ(-\infty ,-1)\cup (\frac{2}{3},\infty )$