<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] First state the conditions that $x$ has to satisfy:


*

*The initial equation requires $x\ge 0$.

*$\sqrt{3+x}=x^2-3$ requires $x^2\ge 3$, i.e., taking into account the previous condition, $x\ge \sqrt3$.
The resulting equation is easy to factorise: rewrite it as
\begin{align}
x^4-x-6x^2+6&=x(x^3-1)-6(x^2-1)=(x-1)\bigl(x(x^2+x+1)-6(x+1)\bigr)\\&=(x-1)(x^3+x^2-5x-6).
\end{align}
One tests the existence of rational roots for the second factor, among $\pm1, \pm2,\pm3,\pm6$, one finds $-2$, so dividing by $x+2$:
$$x^4-x-6x^2+6=(x-1)(x+2)(x^2-x-3).$$
Set $p(x)=x^2-x-3$. Its discriminant is $\Delta=13$. As $p(\sqrt 3)=-\sqrt 3<0$, $\sqrt 3$ separates the roots, so the only root $\ge\sqrt3$ is
$$x=\frac{1+\sqrt{13}}2.$$

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