Simplifying $f(\sqrt{7})$, where $f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}$ 
If $f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}$ ; then $f(\sqrt {7})=\; ?$

I tried solving this equation through many methods, I tried rationalizing, squaring, etc. But after each of them, the method became really lengthy and ugly.
I also noted that once we substitute $\sqrt {7}$ the inner part of the radical becomes imaginary. How to proceed with this piece of information?
Please help me with this problem. Any more innovative methods would be appreciated.

Answer: $4$.

Edit $[7^{th}$ March, $2021]$:
I was told yesterday that this question was wrong and that we were supposed to find $f(2\sqrt5)$. Although I can solve with $2\sqrt5$ the same way as given below in the answers.... What I don't understand is that why $\sqrt7$ doesn't work. Can someone please help?

Using the denesting formula
$$\sqrt{a\pm\sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}} \pm \sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$
We have
$$f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}} \\= 2\sqrt{\frac{x+\sqrt{x^2-16(x-4)}}{2}}=2\sqrt{\frac{x+\sqrt{(8-x)^2}}{2}}$$
Now $\sqrt7<8$, therefore
$$f(\sqrt 7)=2\sqrt{\frac{\sqrt 7+8- \sqrt 7}{2}}=2\cdot 2=4.$$
However as one of the members pointed out, the square root of a nonreal complex number is not well defined. We may get around this by requiring that the "square root" have nonnegative real part, but this is not standard practice.
Update: You can just square both sides directly:
$$(f(x))^2=2x+2\sqrt{x^2-(4-x)^2} = 2x+2\sqrt{(8-x)^2}$$
So if $x\leq 8$ then $f(x)=\sqrt{2x+2(8-x)}=4.$
