<s>[INST] Solve for $x$ in $\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$ 
Solve for $x$:
$$\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$$

I tried to substitute $y=x+2$ and then I try to solve the equation by again and again squaring.
Then I got equation, $$(y-2)(3y^{14}-(y-2)^{15})=0$$
One solution is $y = 2$ and another is $y = 5.$ (I found $5$ as a solution of the equation by hit and trial method).
Therefore, $x = 0$ or $3.$
I'm wondering if there's any  another method to solve it as the repeated squaring step seems to be somewhat absurd.
 [/INST] We want to solve: $\ \sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=x    $
So, we want to solve: $\ \sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2x}}}}=x    \quad \quad (E)$

*

*We see that if $\sqrt{x+2x} =x$, then, $x$ is a solution of $(E)$.
Therefore, $0$ and $3$ are two solutions of $(E)$.

*Suppose that $x$ is a solution of $(E)$ and $x\neq 0$. Then $x>0$ and $F(x)=1$ with
$F(x)=\sqrt{\dfrac{1}{x}+\dfrac{2}{x}\sqrt{\dfrac{1}{x}+\dfrac{2}{x}\sqrt{ \dfrac{1}{x}+\dfrac{2}{x}\sqrt{\dfrac{1}{x}+\dfrac{2}{x}}}}}$

*$F$ is strictly decreasing on $(0,+\infty)$, so, the equation $F(x)=1$ has at most one solution.

We can conclude that $(E)$ has exactly two solutions, $0$ and $3$.
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