Why doesn't my solution give the right answer

Question

I'm trying to solve this inequality:

$1-\frac{2}{\left| x\right|}\leqslant \frac{23}{x^{2}}$

Let's imagine we already know the answer: $x ∈ [-1-2\,\sqrt{6}; 0)⋃(0; 1+2\,\sqrt{6}] $

Now I am going to present my solution (wrong solution):

$1-\frac{2}{\left| x\right|}\leqslant \frac{23}{x^{2}}$ ; $1 = \frac{x^{2}}{x^{2}}$ $(1)$

$\frac{23-x^{2}}{x^{2}}+\frac{2}{\left| x\right|}\geqslant 0$ $(2)$

And now I will multiply the both sides by adjoint expression:

$\left(\frac{23-x^{2}}{x^{2}}+\frac{2}{\left| x\right|}\right)\cdot \left(\frac{23-x^{2}}{x^{2}}-\frac{2}{\left| x\right|}\right)\geqslant 0$ $(3)$

Using $\left(a-b\right)\,\left(a+b\right)=a^{2}-b^{2}$ and the fact that $\left| x\right|^{2}=x^{2}$ I will proceed with my solution.

I stopped on the 3rd step because my solution is already incorrect on it. I am curious why.

Answer 1

As you noticed $|x|^2 =x^2$ so why not use that. Let $t=|x|>0$, since $x=0$ is not a solution. Then we have to solve $$1-{2\over t}\le {23\over t^2}$$ and thus $$ t^2-2t+1\le 24\implies 0<t \le 2\sqrt{6}+1$$ and the rest is easy...