(Infinite) Nested radical equation, how to get the right solution? I've been tasked with coming up with exam questions for a high school math contest to be hosted at my university. I offer the following equation,
$$\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\cdots}}}}=2$$
and ask for the solution for $x$.
Here's what I attempted so far. The first utilizes some pattern recognition, but it gives me two solutions (only one of which is correct).
$$\begin{align*}
\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\cdots}}}}&=2\\
\sqrt{x-\sqrt{x+\sqrt{x-\cdots}}}&=4-x\\
\sqrt{x+\sqrt{x-\cdots}}&=x-(4-x)^2\\
2&=x-(4-x)^2&\text{(from line 1)}\\
(x-6)(x-3)&=0
\end{align*}$$
$x=6$ is the extraneous solution. Where did I go wrong, and how can I fix this?
I know there's a closed form for non-alternating nested radicals $\sqrt{n+\sqrt{n+\cdots}}$ and $\sqrt{n-\sqrt{n-\cdots}}$, but I can't seem to find anything on alternating signs.

You know that:
$$\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\cdots}}}}=A = 2$$
and hence
$$A = \sqrt{x+\sqrt{x-A}} = 2$$
or equivalently
$$\sqrt{x+\sqrt{x-2}} = 2$$
Clearly, $\sqrt{x-2}$ is well defined when $$x \geq 2. ~~~(1)$$
Then, squaring both side, you get:
$$x+ \sqrt{x-2} = 4 \Rightarrow \sqrt{x-2} = 4-x ~~~(2).$$
Since $\sqrt{x-2} \geq 0$, then also $4-x \geq0$, and hence
$$x \leq 4. ~~~(3)$$ 
Joining conditions $(1)$ and $(3)$, one obtain the existence set for $x$:
$$x \geq 2 \wedge x \leq 4 \Rightarrow 2\leq x \leq 4. ~~~(4)$$
Going back to $(3)$, we can square both side and we get:
$$x-2 = (4-x)^2 \Rightarrow x-2=16+x^2-8x \Rightarrow x^2-9x+18=0 \Rightarrow $$ $$\Rightarrow (x-3)(x-6) = 0. ~~~(5)$$
The solution of $(5)$ are $x_1 = 3$ and $x_2 = 6$, but according to $(4)$, only $x_1 = 3$ is feasible.
