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?
 A: 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.
A: 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 $f(x)=4$.
A: 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 $f(x)=\sqrt{2x+2(8-x)}=4.$
A: To answer your latest edit, if a complex number $x$ is not nonnegative real, there are two cases with regard to its square root:

*

*$x<0$. In this case what I have learned is $\sqrt x=\sqrt{-x} i$. Maybe some math textbooks will consider this illegal but many authors comfortably regard $\sqrt{-1}$ as $i$.


*$x=a+bi$ where $b\ne0$. In this case there are two numbers whose square equal to $x$ and one should avoid writing $\sqrt x$.
If $x=\sqrt 7$, it is ok to ask: find all complex numbers $z$ such that $z^2=\sqrt7 - 4\sqrt {\sqrt 7 - 4}$. It's also ok to ask you to find all complex numbers $z$ such that $z^2=\sqrt 7 + 4\sqrt {\sqrt 7 -4}$. Each will have two solutions (one being the opposite of the other). If you add the two (separate) solutions with positive real part you will get $4$. If you add the two solutions with negative real part you will get $-4$.
