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?

Let $f(x) = 4 g(t)$, $x = 4(t^2+1)$. Then, provided that $\Re t \notin\{ 1, -1\}$,
$$\begin{split}
  g(t) &= \tfrac{1}{2}\left(
    \sqrt{(1-t)^2}+\sqrt{(1+t)^2}\right) \\
    &= \begin{cases}
    -t & \Re t < -1 \\
    1 & -1 < \Re t < 1 \\
    t & \Re t > 1
    \end{cases}\text{.}
\end{split}$$
That expression is found by the complex-number identity
$$\sqrt{a^2}=\begin{cases}
  a & \Re a > 0 \\
  -a & \Re a < 0
\end{cases}\text{,}$$
and the constraint on the domain comes from ignoring the branch cut. Then when $x=\sqrt{7}$, $t=\pm \mathrm{i}\sqrt{1-\tfrac{\sqrt{7}}{4}}$, so $g(t)=1$ and $f(x)=4$.
