Find the value of $x$ such that $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$ Find the value of $x$, $$\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$$
Help guys please, I have tried and I got, $x=-2, x=1$, and I think it's wrong
 A: $$\begin{align}\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}&=x&\Longleftrightarrow \\
\left(\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}\right)^2&=x^2&\Longleftrightarrow\\
4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}&=x^2&\Longleftrightarrow\\
\left(\sqrt{4-\sqrt{4+\sqrt{4-x}}}\right)^2&=\left(x^2-4\right)^2&\Longleftrightarrow\\
4-\sqrt{4+\sqrt{4-x}}&=\left(x^2-4\right)^2&\Longleftrightarrow\\
-\sqrt{4+\sqrt{4-x}}&=\left(x^2-4\right)^2-4&\Longleftrightarrow\\
\sqrt{4+\sqrt{4-x}}&=-\left(x^2-4\right)^2+4&\Longleftrightarrow\\
\left(\sqrt{4+\sqrt{4-x}}\right)^2&=\left(-\left(x^2-4\right)^2+4\right)^2&\Longleftrightarrow\\
4+\sqrt{4-x}&=\left(-\left(x^2-4\right)^2+4\right)^2&\Longleftrightarrow\\
\sqrt{4-x}&=\left(x^4-8x^2+12\right)^2-4&\Longleftrightarrow\\
\left(\sqrt{4-x}\right)^2&=\left(\left(x^4-8x^2+12\right)^2-4\right)^2&\Longleftrightarrow\\
4-x &=\left(x^4-8x^2+10\right)^2\left(x^4-8x^2+14\right)^2 &&
\end{align}
$$
Solving $x$ gives us $\frac{1}{2}+\frac{\sqrt{13}}{2}$
A: Substitute back the expression for x on the left, and repeat.  You arrive, eventually, with the infinite expression $$ x = \sqrt{4+\sqrt{4-\sqrt{4+...}}}$$
By symmetry we are inclined to define $$ y = \sqrt{4-\sqrt{4+\sqrt{4-...}}}$$
Ignoring convergence issues (settled by Herschfeld..I believe Poyla has a good proof, but don't have an exact reference at hand) we easily see that $$x^2 = 4+y$$ and $$y^2 = 4 - x$$
This system is easily solved numerically (I didn't try any other means) and we get $x = 2.30277...$ and it is easy to check that this value works.
A: HINT:$$f(\bar{x})=\bar{x}$$
EDIT
$$f(\bar{x})=x$$ $$\bar{x}=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}$$
start putting $x=0$
now you have $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-0}}}}\approx 2.29023...$
now do the same things with the previous result and you get:
$$\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-2.29023}}}}\approx 2.3026...$$
If you want a better approximation do it again (you'll get $2.302775...$)
A: The polynomial generated by @JanEerland's method will yield a bunch of false roots because every time you square you loose sign information.  So you need to test all 16 suggested roots with the original equation.
Here's my Mathematica code:
f[x_] := x^2 - 4;
eqn = f[f[f[f[x]]]] + x // Expand;
aa = Solve[eqn == 0, x];

Output (abbreviated):
(*{{x -> 1/2 (1 - Sqrt[13])}, 
{x -> 1/2 (1 + Sqrt[13])}, 
{x -> 1/2 (-1 - Sqrt[17])}, 
{x -> 1/2 (-1 + Sqrt[17])}...   *)

Now test this solution set against the original equation:
Sqrt[4 + Sqrt[4 - Sqrt[4 + Sqrt[4 - x]]]] - x /. aa // N

Output:
    (*  {3.5876, 0., 4.84176, 0.736553, 4.78663, 4.66183, 4.56239, 3.95255, \
3.81752, 3.48907, 1.08211, 1.01294, 0.650461, -0.0492326, -0.22168, \
-0.246164}  *)
Only the second root in the list returns a zero, so the answer is 1/2 (1 + Sqrt[13])
A: This should be a comment but then too long ...
Picking up on the solution given by Jan Erlang: 
$$\left(x^4-8 x^2+10\right)^2 \left(x^4-8 x^2+14\right)^2+x-4=\left(x^2-x-3\right) \left(x^2+x-4\right) \left(x^{12}-24
   x^{10}-x^9+228 x^8+16 x^7-1087 x^6-88 x^5+2720 x^4+191 x^3-3380
   x^2-136 x+1633\right)$$
Now $x^2-x-3=0$ yields: $\frac{1}{2} \left(1\pm\sqrt{13}\right)$ and $x^2+x-4=0$ yields $\frac{1}{2} \left(-1\pm\sqrt{17}\right)$ (where only positive roots are acceptable). Further $$x^{12}-24
   x^{10}-x^9+228 x^8+16 x^7-1087 x^6-88 x^5+2720 x^4+191 x^3-3380
   x^2-136 x+1633=0$$ yields $6$ more positive roots.
A: We can solve this problem completely, without looking at anything larger than a quartic polynomial.
Let $f(x)=\sqrt{4+\sqrt{4-x}}$.  The equation we are trying to solve is $f(f(x))=x$.  
Observe that $f(x)$ is strictly decreasing from $f(0)=\sqrt6\gt0$ to $f(4)=2\lt4$, which means the equation $f(x)=x$ has a unique solution, which is relatively easy to find:
$$\begin{align}
x=\sqrt{4+\sqrt{4-x}}
&\implies x^2=4+\sqrt{4-x}\\
&\implies (x^2-4)^2=4-x\\
&\implies x^4-8x^2+x-12=0\\
&\implies(x^2-x-3)(x^2+x-4)=0
\end{align}$$
There are two roots that satisfy $0\le x\le4$:
$$x={1+\sqrt{13}\over2}\quad\text{and}\quad x={\sqrt{17}-1\over2}$$
Only the first of these satisfies the equation $x=\sqrt{4+\sqrt{4-x}}$; the other one, $x=(\sqrt{17}-1)/2$, satisfies the equation $x=\sqrt{4-\sqrt{4-x}}$.
Since $x=(1+\sqrt{13})/2$ satisfies $f(x)=x$, it also satisfies $f(f(x))=x$, so we have found $a$ solution of our equation.  To see that there are no other solutions, we can appeal to calculus.
From $f(x)=\sqrt{4+\sqrt{4-x}}$ we have
$$f'(x)={-1\over4\sqrt{4+\sqrt{4-x}}\sqrt{4-x}}$$
from which it will turn out to be convenient to note that, for $x\le3$, $|f'(x)|\le|f'(3))|={1\over4\sqrt5}\lt1$
Meanwhile, since $f(x)$ is decreasing on $0\le x\le4$, the function $F(x)=f(f(x))$ is increasing, hence
$$2\lt\sqrt{4+\sqrt{4-\sqrt6}}=f(f(0))\le f(f(x))\le f(f(4))=\sqrt{4+\sqrt2}\lt3$$
(The crude outer bounds, $2$ and $3$, are just there to simplify some of the arithmetic we're about to do.)  Thus any solution to $x=f(f(x))$ satisfies $2\lt x\lt3$, which means $f(x)\lt f(2)=\sqrt{4+\sqrt2}\lt3$.
Now suppose the equation $f(f(x))=x$ had two solutions, say $x=a$ and $x=b$ with $a\lt b$.  Then by the Mean Value Theorem applied to $F(x)=f(f(x))$, $F'(c)=1$ for some $c\in(a,b)$.  But $2\lt a\lt c\lt b\lt 3$ implies $f(c)\lt3$, and we thus see that
$$|F'(c)|=|f'(f(c))f'(c)|\lt|f'(3)|^2\lt1$$
which is a contradiction.  We thus conclude that $x={1+\sqrt{13}\over2}$ is the only solution to the equation $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$
