Solve the inequality $\frac{1}{|x|}-x >2$ Solve the inequality $\frac{1}{|x|}-x >2$ 
My attempt: 
$|x|=x, x>0$ 
$|x|=-x, x<0$ 
$1)$ for $x>0$ 
$\frac{1}{x}-x>2$ 
$-x^{2}-2x+1>0 \Rightarrow x \in(-1-\sqrt 2, -1+\sqrt 2)$ but since $x>0$ then $x\in (0, -1+\sqrt 2).$ 
$2)$ if $x<0$ 
$\frac{-1}{x}-x>2$ 
$\frac{-1-x^{2}-2x}{x}>0 \Rightarrow -x^{2}-2x-1<0$ 
$\Rightarrow x\in \Bbb R \setminus0$ but since $x<0 \Rightarrow x\in (-\infty, 0).$
So my final solution is $x\in(-\infty,0)\cup(0,\sqrt 2 -1)$ 
But the solution should be $x\in(-\infty,-1)\cup(-1,0)\cup(0, \sqrt 2 -1)$. So I should exclude $-1$ from my solution, but I can't see where  I missed that step. 
 A: In the negative case you reached the statement that $-x^2-2x-1<0$ but you didn't finish looking at it.
$$-x^2-2x-1<0$$
$$x^2+2x+1>0$$
$$(x+1)^2>0$$
$$x\neq-1$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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With $\ds{x \not= 0}$:

\begin{align}
&{1 \over \verts{x}} - x > 2 \implies
1 - \mrm{sgn}\pars{x}x^{2} > 2\,\mrm{sgn}\pars{x}x \implies
x^{2} + 2x - \mrm{sgn}\pars{x} \begin{array}{c}< \\[-2mm] >\end{array}\ 0
\\ &
\pars{~<,\ >\ \mbox{correspond to}\ \,\mrm{sgn}\pars{x} = \pm\, 1,\ \mbox{respectively}~}.
\\[5mm] &
x_{\pm} = -1 \pm \root{1 + \mrm{sgn}\pars{x}}\quad
\mbox{are the roots of}\quad
x^{2} + 2x - \mrm{sgn}\pars{x} = 0
\end{align}



*

*$\ds{\color{#f00}{\mrm{sgn}\pars{x} < 0} \implies x_{\pm} = -1}$:
$$
x^{2} + 2x + 1 > 0\implies \pars{~\color{#f00}{x < 0}\quad \mbox{because}\quad
\mrm{sgn}\pars{x} < 0 ~}
$$



*$\ds{\color{#f00}{\mrm{sgn}\pars{x} > 0} \implies x_{\pm} = -1 \pm \root{2}}$:
$$
x^{2} + 2x - 1 < 0\implies \pars{~\color{#f00}{0 < x < \root{2} - 1}\quad \mbox{because}\quad
\mrm{sgn}\pars{x} > 0 ~}
$$

