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 ?
 A: 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).
A: $$\sqrt{3+x}=x^2-3$$
$$3+x=(x^2-3)^2$$
$$x^4 -6x^2 +9 -x - 3=x^4 -6x^2-x+6=0$$
$$(x-1)(x+2)(x^2-x-3)=0$$
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 $x=\frac{1+\sqrt{13}}{2}$
A: 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
$$x=\frac{1+\sqrt{13}}2.$$

A: We have to factorize $x^4 - 6x^2 - x + 6 = 0.$
We would observe that the values of x would be $1,-2,\frac{1±\sqrt{13}}{2}$.
