Solving for $x$ when $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x$ Suppose $x$ is a real number such that $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x.$ Find $x.$

I first squared to get rid of the first square root, which gave me $\sqrt{3} - \sqrt{\sqrt{3} + x} = x^2.$ However, I'm not sure how to move on from there. Can someone give me a hint?
 A: From where you left off:
$$\begin{align}
\sqrt{3}-\sqrt{\sqrt3+x}
&=x^2\\
\sqrt{3}-x^2
&=\sqrt{\sqrt3+x}\\
3-2\sqrt{3}x^2+x^4
&=\sqrt{3}+x\\
x^4-2\sqrt{3}x^2-x+\left(3-\sqrt3\right)&=0\\
\left(x^2+x+(1-\sqrt{3})\right)\left(x^2-x-\sqrt{3}\right)&=0
\end{align}$$
So if there is a solution to the original equation, it is a root of this 4th degree polynomial. It's easy to find its four roots since it factors. But some roots of this polynomial might not solve the original equation, since we squared a few times earlier. So each one should be checked.
Note that the original left-hand side is not real unless $x$ is in $\left[-\sqrt{3},3-\sqrt{3}\right]$. That should help eliminate several of the four polynomial roots.
A: Let $a = \sqrt{3}$, so $x = \sqrt{a-\sqrt{a+x}}$. Let's call this equation $(*)$.
Let $y = \sqrt{a+x}$ so $x = \sqrt{a-y}$. Therefore

*

*$x^2 = a-y$

*$y^2 = a+x$
Subtracting 2. from 1. we have $x^2 - y^2 = -(x+y)$. Let's consider the following cases:

*

*If $x+y = 0$, then $x = -\sqrt{a+x}\le 0$, but from $(*)$ we have $x\ge 0$, so we must have $x = 0$. It's easy to see that this doesn't work.


*If $x+y \neq 0$, we have $\begin{aligned}\frac{x^2-y^2}{x+y} = -1 \therefore x-y = -1\end{aligned}$, i.e., $x+1 = \sqrt{a+x}$, so $(x+1)^2 = x^2+2x+1 = a+x$, therefore $x^2+x+1-a = 0$, which can be solved as a quadratic equation to get $\begin{aligned}x = \frac{-1+ \sqrt{1-4(1-a)}}{2} = \frac{-1+\sqrt{4a-3}}{2}\end{aligned}$ (notice that this is the only positive solution).
A: Comment:
If you put $x=\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} $ under the third radical you get:
$$\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + \sqrt{\sqrt{3} - \sqrt{\sqrt{3} + \sqrt{\sqrt{3} - \sqrt{\sqrt{3} + }} . . .}} }} = x$$
Suppose there is no negative sign then by squaring both sides you get:
$$X^2=\sqrt 3 +X$$
Which gives:
$$X=\frac{-1 ±\sqrt{1+4\sqrt 3}}{2}≈0.9≈\frac{\sqrt3}{2}$$
So $x<X≈\frac{\sqrt 3}2$
