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.

Let $$f(x)=\sqrt{2+\sqrt{2-x}}$$ and
$$g(x)=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$$
See the fig below where $y=f(x)$ blue and $y=g(x)$ green and $y=x$ (red).
The solution of $x=f(x)$ is given by the intersection of red and blue lines at $x=1.6180...=\frac{1+\sqrt{5}}{2}$ (the same golden number)
The root of $x=g(x)$ is also given intersection of red and  green  at  the golden number.
But the roots of the equation $f(x)=g(x)$ is given by two points of intersection of blue and green lines, one at the golden number and the other at $x=-0.8959...$
Also it turns out that the common root of $x=f(x), x=g(x), f(x)=g(x)$ is at $\frac{1+\sqrt{5}}{2}$ and not at $\frac{\sqrt{5}-1}{2}.$

