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?

Set $\sqrt{x-4}=u \to x=u^2+4\\$so
$$\sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}=\\ \sqrt{u^2+4-4u}+\sqrt{u^2+4+4u}=\\
|u-2|+|u+2|=\\
|\sqrt{x-4}-2|+|\sqrt{x-4}+2|$$ but when deals to imaginary numbers absolute sign is not necessary.
