<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] $$\sqrt{3+x}=x^2-3$$
$$3+x=(x^2-3)^2$$
$$x^4 -6x^2 +9 -x - 3=x^4 -6x^2-x+6=0$$
$$(x-1)(x+2)(x^2-x-3)=0$$
None of $1, -2, \frac{1-\sqrt{13}}{2}$ are not answers, due to our double-squaring. Basically, we need the following inequalities to be true:
$$x+3 \ge 0$$
$$x\ge 0$$
$$\sqrt{3+x} = x^2 - 3 \ge 0$$
Thus $x \ge \sqrt{3}$, and $x=\frac{1+\sqrt{13}}{2}$
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