Solutions of the equation $x=\sqrt{2 + \sqrt{2 +\sqrt 2+.....}}$ I have seen the equation $x=\sqrt{2 + \sqrt{2 +\sqrt 2+.....}}$ in many places and the answer is $x=2$ which is obtained by making the substitution for $x$ inside the radical sign i.e $x=\sqrt{2+x}$ which renders the quadratic $x^2 -x -2=0$ 
having the solutions $x=-1$ and $x=2$ and we just ignore $x=-1$ since it is negative. 
But my question is, we could also have made the substitution for $x$ under the second radical sign i.e. $x=\sqrt{2 + \sqrt{2 +x}}$ and get the bi-quadratic $x^4-4x^2-x+2=0$. Which has the solutions 
$$x=-1,2, \frac{-1-\sqrt{5}}{2} , \frac{-1+\sqrt{5}}{2}$$ Or we could also have substituted anywhere else in the infinite radical expression and it would have given us some real roots. So how do we decide that $x=2$ is the correct solution?
 A: You can obtain $x$ using the recursive sequence 
$$\begin{cases} x_0 = 0\\ x_{n+1}=\sqrt{2+x_n}\end{cases}$$
Since both $x_n$ and $x_{n+1}$ converge to $x$ and $g(t)=\sqrt{2+t}$ is a continuous function, one must have
$$
\lim x_{n+1} = \sqrt{2+\lim x_n} \Rightarrow x^2 =2+x.
$$
Note: You can check convergence using the fixed point theorem.In fact, $g$ is invariant and contractive in $[0,+\infty[$ and you can conclude that the sequence converges to a positive fixed point of $g$, for any choice of $x_0 \ge 0$.
A: Note that if that expression defines a number $x$, then $x$ satisfies $x^2-x-2$. That $x$ satisfies some other relations is not relevant. That some other numbers satisfy those other relations is even less relevant; those other numbers do not satisfy $x^2-x-2$.
As an aside; to prove that $x=2$ it is also necessary to prove that
$$x=\sqrt{2 + \sqrt{2 +\sqrt{2+.....}}},$$
actually defines a number. This comes down to showing that a certain infinite sequence converges.
A: For $x=\sqrt{2 + \sqrt{2 +x}}$ you got the bi-quadratic $x^4-4x^2-x+2=0$ with solutions 
$$x=-1,2, \frac{-1-\sqrt{5}}{2} , \frac{-1+\sqrt{5}}{2}.$$
Now, the first and third values ($x=-1, \frac{-1-\sqrt{5}}{2}$) can not be a solution because they are negative.
The fourth value is approximately $0.61803398875,$ but we know that The value of $x$ has to be greater than $\sqrt 2$. So, this can not be a solution as well.
After eliminating all the other possibilities, $x=2$ is the solution.
If you make the substitution for $x$ under the $n$-th radical sign, you get an equation with a degree $2^n$.
We do not have a method (It cannot be solved algebraically in terms of a finite number of additions, subtractions, multiplications, divisions, and root extractions) to solve equations of degree $> 5$.
