Solve the inequality $\frac{1}{|x|}-x >2$

Question

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.

Answer 1

In the negative case you reached the statement that $-x^2-2x-1<0$ but you didn't finish looking at it.

$$-x^2-2x-1<0$$

$$x^2+2x+1>0$$

$$(x+1)^2>0$$

$$x\neq-1$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}

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

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