<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 have,
$$\begin{align}&\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x≥0\\
\iff &2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=2x\\
\iff &x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=3x\\
\iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}=\sqrt{3x}\\
\iff &2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}=2\sqrt{3x}\\
\iff &x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}=x+2\sqrt{3x}\\ 
\iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}} = \sqrt{x+2\sqrt{3x}}\end{align}$$
Let, $\sqrt{x+2\sqrt{3x}}=u,\thinspace u≥0$ then we get,
$$\begin{align}\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2u}}}}=u\end{align}$$
Then, we see that,
$$\begin{align}&\sqrt{x+2u} = u\\
\iff &\sqrt{x+2\sqrt{x+2u}} = u\\
\iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2u}}}=u\\
\iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2u}}}}= u\end{align}$$
Therefore, we have
$$\begin{align}&x+2u= x+2\sqrt{3x}\\
\iff &u^2=3x\\
\iff &x+2\sqrt{3x}=3x\\
\iff &\sqrt{3x}=x\\
\iff &x\in\left\{0,3\right\}.\end{align}$$
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