<s>[INST] Solve $\sqrt{3+\sqrt{3+x}}=x$ 
Solve
  $$\sqrt{3+\sqrt{3+x}}=x$$

My try:
$$\sqrt{3+\sqrt{3+x}}=x \\ 
3+\sqrt{3+x}=x^2\\\sqrt{3+x}=x^2-3\\3+x=(x^2-3)^2$$
$$x^4-6x^2+9=x+3\\x^4-6x^2-x+6=0$$
Now ?
 [/INST] There is no need to deal with any quartic equation in $x$. The key is the function
on LHS $x \mapsto \sqrt{ 3 + \sqrt{3+x}}$ is a composition of the map 
$x \mapsto \sqrt{3+x}$ with itself.
The map $x \mapsto \sqrt{3+x}$ is strictly increasing whenever it is defined (i.e $x \ge -3$)
If $\sqrt{3+x} > x$, then $\sqrt{3 + \sqrt{3+x}} > \sqrt{3+x} > x$.
If $\sqrt{3+x} < x$, then $\sqrt{3 + \sqrt{3+x}} < \sqrt{3+x} < x$.
If we want $\sqrt{3 + \sqrt{3+x}} = x$, we need  $\sqrt{3+x} = x$. This leads to
$$x^2 = 3 + x \iff x^2- x - 3 = 0 \implies x = \frac{1 \pm \sqrt{13}}{2}$$
Since $x = \sqrt{3 + \sqrt{3 + x}}$ is supposed to be non-negative, 
$x = \frac{1 + \sqrt{13}}{2}$ is the only possible solution (and it does work).
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