(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.
 A: 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.
A: The problem with that kind of expression with nested radicals is that the rigorous meaning of the "$\ldots$" at the end is not always clear (and I presume that's what some people were trying to point out in the comments). 
In this case, I think you are considering the sequence given by $x_1 = \sqrt{x}$ and $x_{n+1} = \sqrt{x + \sqrt{x - x_n}}$ (for some fixed $x \geq 0$) and asking for values of $x$ for which $\lim_{n \to \infty} x_n = 2$.
The first thing to be worried about here is whether it's safe to take $\sqrt{x - x_n}$, that is, whether $x_n \leq x$. For $0 < x < 1$ everything goes south as $x_1 = \sqrt{x} > x$. For $x > 1$, $x_1 = \sqrt{x}$ is ok, but $x_2 = \sqrt{x + \sqrt{x - \sqrt{x}}}$ can be a problem if $x$ is small (try $x = 1.1$); to avoid these issues, let us assume $x \geq 2$, when clearly $x_n \leq \sqrt{x + \sqrt{x}} < x$.
Secondly, does $\lim_{n \to \infty} x_n$ always exist if $x \geq 2$? Yes it does, because the sequence is clearly increasing (why?) and bounded above by $\sqrt{x + \sqrt{x}}$ (why?).
Since the limit exists, we can call it $L = L(x)$: it satisfies the inequality $\sqrt{x} \leq L \leq \sqrt{x + \sqrt{x}}$ and the equation $L = \sqrt{x + \sqrt{x - L}}$, so that $L^4 - 2L^2x + L + x^2 - x = 0$.
Which values of $x$ could possibly have $L(x) = 2$? Well, they must satisfy $x^2 - 9x + 18 = 0$, so that $x = 3$ or $x = 6$. On the other hand, since $x$ also must satisfy $\sqrt{x} \leq 2$ (or $x \leq 4$), the solution $x = 6$ can be discarded.
What about $x = 3$, can it be proclaimed "the winner"? Not quite yet, we must verify that $L(3) = 2$: for that, we check that there are no other viable candidates for $L(3)$. To see this, notice that $L(3)$ must be a root of
$L^4 - 6L^2 + L + 6 = (L+1)(L-2)(L^2 + L - 3)$. Since $2$ is the only root satisfying $\sqrt{3} \leq 2 \leq \sqrt{3 + \sqrt{3}}$, we can safely say $L(3) = 2$, and $x = 3$ is the only solution to the original problem.
A: In the second line, ......$=4-x$. The LHS is a square root which is positive. So $x<4$. After you take the square $(4-x)^2$, you introduced an extra root, which should be rejected at the end.
