Equality of two iterated square roots Solve for $x$:
$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}}=\sqrt{x\sqrt{x\sqrt{x\sqrt{x\dots}}}}$
My attempt:
The L.H.S is equal to $\dfrac{1+\sqrt{4x+1}}{2}$ and R.H.S equals $x^2$
Equating both sides:
$\implies 4x+1=(2x^2-1)^2$
$\implies 4x+1=4x^4-4x^2+1$
$\implies 4x^4-4x^2-4x=0$
$\implies x(x^3-x-1)=0$
Disregarding the complex roots,
$\implies x=0$ 0r $\dfrac{1}{3}\sqrt[3]{\dfrac{27-3\sqrt{69}}{2}}+\dfrac{\sqrt[3]{\dfrac{9+\sqrt{69}}{2}}}{3^{2/3}}$
Is my solution correct? By the way I would like to see other methods to solve it. Thanks!

I get a different solution:
Let $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+..}}}}=\sqrt{x\sqrt{x\sqrt{x\sqrt{x...}}}}$.
Then $y^2 = x+ y$ and $y^2 = xy$.
From the second equation we see that $x = y = 0$ is a possible solution, otherwise, $x = y \neq 0$.
So from the first equation: $x^2 = 2x \Rightarrow x = y =  2$
