A golden question $\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$ How would you solve this problem for real $x$?
$$\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$$

It can be easily shown that both equations
$$x=\sqrt{2+\sqrt{2-x}}\tag{1}$$
and $$x=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}\tag{2}$$
have the same real solution, i.e. the golden ratio, $$x = \frac {\sqrt 5 + 1}2=\phi.$$
The main question is derived by combining (1) and (2) into one equation. The final solution is still the same, but the approach might not be as straightforward.

With some difficulty.
You can turn the equation into a polynomial through lots of squaring and rearranging, starting with:
$\begin{eqnarray} \sqrt{2+\sqrt{2-x}} & = & \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} \\
& = & \sqrt{\frac{x - 1}{x}}\left(\sqrt{x + 1} + 1\right) \\
2 + \sqrt{2 - x} & = & \frac{x - 1}{x} \left( x + 2 + 2 \sqrt{x + 1} \right) 
 \\
\frac{(x-1)(x+2)}{x} - 2 & = & \sqrt{2 - x} - 2 \frac{x - 1}{x}\sqrt{x + 1} \\
x^2 - x - 2 & = & x\sqrt{2 - x} - 2(x-1)\sqrt{x+1} \end{eqnarray}$
After a few more rounds, you should get to something vaguely resembling $x^8-10x^7+39x^6-22x^5-63x^4+32x^3+32x^2 = 0$. You can divide out $x^2$ (which is clearly not a solution to the original equation), and then you can use the knowledge that $\varphi$ is one of the roots to divide out the quadratic $x^2 - x - 1$ (since the polynomial has integer coefficients, you can deduce that $-\varphi^{-1}$ is also a root, which is an erroneous solution introduced in one of the squaring steps).
At this point, you're left with $x^4-9x^3+31x^2-32 = 0$. This quartic has two complex roots and two real roots, but actually figuring them out analytically is what mathematicians call "damn hard", and is commonly achieved via the method known as "chuck it into Mathematica".
