<s>[INST] 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
 [/INST] 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 $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$
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