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?

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.
