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

Question

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)

So what is it you want to know? How to solve this problem quickly? Or how to proceed with your method? — Panglossian Oporopolist, Jul 17 '15 at 02:49

score 4 · Answer 1 · answered Jul 17 '15 at 02:51

4

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.

answered Jul 17 '15 at 02:51

Dr. Sonnhard Graubner

95,283

@ClaudeLeibovici Squares are always bigger than zero, so $13x^6+117x^4+78x^2+1>0$ always. – wythagoras Jul 17 '15 at 06:52
$\left(9\sqrt{9x^2-4}\right)^2=\left(\frac{2}{x}+\frac{18x}{x^2+1}-25x\right)^2$ – user236182 Jul 17 '15 at 10:09

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

1 Answers1