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!

Note that RHS is wrong:
$$y=\sqrt{x\sqrt {x\cdots}}\implies y^2=x\sqrt{x\sqrt {x\cdots}}=xy\implies y=0 \text{ or }x$$
