Equality of two iterated square roots

Question

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!

Answer 1

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: $x^2 = 2x \Rightarrow x = y = 2$

Answer 2

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