Equation with huge number of nested square roots:

Hint: $\displaystyle\;\; a_n-a_{n-1}=\sqrt{x+2a_{n-1}}-\sqrt{x+2a_{n-2}}=\frac{2(a_{n-1}-a_{n-2})}{\sqrt{x+2a_{n-1}}+\sqrt{x+2a_{n-2}}}\,$, so the differences between consecutive terms have the same sign.

Rather than comparing $a_n$ to $a_{n-1}$, it is much simpler to compare $a_n$ to $x$. Observe that if $0<x<3$ and $y\geq x$ then $$\sqrt{x+2y}\geq\sqrt{3x}>\sqrt{x^2}=x.$$ It follows that if $0<x<3$, then $a_n(x)>x$ for all $n>0$ (by induction on $n$). Similarly (just reverse all the inequalities), if $x>3$, then $a_n(x)<x$ for all $n>0$. So the only way we can ever have $a_n(x)=x$ for any $n>0$ is if $x=3$ or $x=0$.