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.
 A: Mark $a= 3x$ and $b= 2x-1$ then we get:
$$a(\sqrt{a^2+3}+2)+b(\sqrt{b^2+3}+2)=0$$
Now the function $f(t) =  t(\sqrt{t^2+3}+2)$ is odd and strictly increasing so it is also injective. Since $f(a)+f(b)=0$ we get $$f(a) = -f(b) = f(-b)\implies a=-b \implies x={1\over 5}$$
A: Hint: Write your equation in the form
$$3x\sqrt{9x^2+3}+(4x-2)\sqrt{x^2-x+1}=2-10x$$ and square both sides.
You will get
$$6x(4x-2)\sqrt{9x^2+3}\sqrt{x^2-x+1}=-x(97x^3-32x^2-37x+20)$$
Squaring one more times we get
$$-{x}^{2} \left( 169\,{x}^{4}+234\,{x}^{3}-695\,{x}^{2}+360\,x-32
 \right)  \left( -1+5\,x \right) ^{2}
=0$$
A: It is a nicely set up problem:
$$(3x)(2+\sqrt{9x^2 + 3}) + (4x-2)(\sqrt{x^2 - x +1}+1) = 0 \iff \\
(6x)(1+\sqrt{\frac94x^2 + \frac34}) = (2-4x)(1+\sqrt{x^2 - x +1}) \\
\begin{cases}6x=2-4x\\ \frac94x^2 + \frac34=x^2-x+1\end{cases} \Rightarrow \begin{cases}x=\frac15\\ 5x^2+4x-1=0 \Rightarrow x=\frac15;-1\end{cases} \Rightarrow x=\frac15.$$
Addendum: The left hand side function is:
$$\begin{cases}f(x)<0,x\le 0\\ f(x)>0,x\ge \frac12\end{cases}\\
f'(x)=3(2+\sqrt{9x^2+3})+3x\cdot \frac{9x}{\sqrt{9x^2+3}}+\\
4(\sqrt{x^2-x+1}+1)+(2x-1)\cdot \frac{2x-1}{\sqrt{x^2-x+1}}>0,0<x<\frac12$$
Hence, the single root is $x=\frac15$.
A: 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!
