<s>[INST] Solve $3x(2+\sqrt{9x^2 + 3}) + (4x-2)(\sqrt{x^2 - x +1}+1) = 0$ for $x\in \mathbb{R}$ 
Solve the equation in $ \mathbb{R}$: $$3x(2+\sqrt{9x^2 + 3}) + (4x-2)(\sqrt{x^2 - x +1}+1) = 0$$ 

I've been tried to solve this question for  3 hours, but can't find out any answers.
Just like I running in the maze, if I represent  $\sqrt{9x^2 + 3} = A$, $ \sqrt{x^{2}-x+1}= B$
Finally we've got $2A^{2} -15 = 18B^2 + 9\sqrt{4B^2 +1}$  , which is not help me to find relation between $A$ and $B$ anymore. 
I just wonder if we have a nice solution approaches to the problem.
I appreciate any suggestion and help. Thank you.
 [/INST] It's obvious that for $x\leq0$ our equation has no real roots and since
$$\left(4x-2)\sqrt{x^2-x+1}\right)'=\frac{8x^2-8x+5}{\sqrt{x^2-x+1}}>0,$$ we see that $$10x-2+3x\sqrt{9x^2+3}+(4x-2)\sqrt{x^2-x+1}$$ increases for $x>0$  and our equation has one real root maximum.
But $\frac{1}{5}$ is a root and we are done!
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