Solve equation: $25x+9\sqrt{9x^2-4}=\frac{2}{x}+\frac{18x}{x^2+1}$ Solve equation: $25x+9\sqrt{9x^2-4}=\dfrac{2}{x}+\dfrac{18x}{x^2+1}$

I used wolframalpha.com and got the only solution $x=-\dfrac{1}{\sqrt2}$
And this is my try:

Domain: $|x|\ge\dfrac{2}{3}$
If $x\ge\dfrac{2}{3}$, we have:
$\dfrac{2}{x}+\dfrac{18x}{x^2+1}\le3+\dfrac{18x}{2x}=12$ and $25x+9\sqrt{9x^2-4}\ge25x\ge\dfrac{50}{3}$   (no solution)
If $x\le\dfrac{-2}{3}$,... (I have no idea in this case)


rewrite your equation in the form
$$9\sqrt{9x^2-4}=\left(\frac{2}{x}+\frac{18x}{x^2+1}-25x\right)^2$$
this can be written in the form
$$4(-1+2x^2)(1+78x^2+117x^4+13x^6)=0$$
solving this we get $$x=\pm\frac{1}{\sqrt{2}}$$
only the number with $$x=-\frac{1}{\sqrt{2}}$$ is the searched solution.
