Solve $3x(2+\sqrt{9x^2 + 3}) + (4x-2)(\sqrt{x^2 - x +1}+1) = 0$ for $x\in \mathbb{R}$

Question

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.

Answer 1

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 $$-{x}^{2} \left( 169\,{x}^{4}+234\,{x}^{3}-695\,{x}^{2}+360\,x-32 \right) \left( -1+5\,x \right) ^{2} =0$$

Answer 2

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!

Answer 3

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 $$f(a) = -f(b) = f(-b)\implies a=-b \implies x={1\over 5}$$

Answer 4

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 $x=\frac15$.