How to solve $x<\frac{1}{x+2}$ Need some help with:
$$x<\frac{1}{x+2}$$
This is what I have done:
$$Domain: x\neq-2$$
$$x(x+2)<1$$
$$x^2+2x-1<0$$
$$x_{1,2} = \frac{-2\pm\sqrt{4+4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{-2\pm2\sqrt{2}}{2}$$
What about now?

Hint 1: You already did a mistake: When you multiply by $(x+2)$ the inequality changes if $x+2<0$.
So first split the problem in two cases: $x+2 >0$ and $x+2<0$.
Hint 2 Factor $x^2+2x-1=(x-x_1)(x-x_2)$ and disscuss the sign of each linear term.
