Assume that we have $x = - (t + 1/t)^2 = -t^2 - 1/t^2 - 2$ for some $t \in \mathbb{C}$. Then $$ \sqrt{-x} = t + \frac{1}{t} = \frac{t^2 + 1}{t},\quad x + 4 = -\left(t - \frac{1}{t}\right)^2, \quad \sqrt{x(x + 4)} = t^2 - \frac{1}{t^2}, \\ x + \sqrt{x(x + 4)} = -2\left(1 + \frac{1}{t^2}\right) = -2\frac{1 + t^2}{t^2}, \quad x - \sqrt{x(x + 4)} = -2(1 + t^2) $$ Consequently, we get $$ f_n(x)= \tilde f_n(t) = \frac{(1 + t^2)^n(t^{n + 2} - 1)^2}{t^{2n}(t^2 - 1)^2},\quad\text{if } n\text{ is even}, \\ f_n(x) = \tilde f_n(t) = -\frac{(1 + t^2)^n(1 - t^{n -1})(1 - t^{n+5})}{t^{2n}(t^2 - 1)^2},\quad\text{if } n\text{ is odd}. $$ Hence the only roots of $\tilde f_n$ are the suitable roots of unity and the roots of $f_n$ are all in the real segment $[0, 4]$ and their closure is the whole segment for large $n$.

Assume that we have $x = - (t + 1/t)^2 = -t^2 - 1/t^2 - 2$ for some $t \in \mathbb{C}$. Then $$ \sqrt{-x} = t + \frac{1}{t} = \frac{t^2 + 1}{t},\quad x + 4 = -\left(t - \frac{1}{t}\right)^2, \quad \sqrt{x(x + 4)} = t^2 - \frac{1}{t^2}, \\ x + \sqrt{x(x + 4)} = -2\left(1 + \frac{1}{t^2}\right) = -2\frac{1 + t^2}{t^2}, \quad x - \sqrt{x(x + 4)} = -2(1 + t^2) $$ Consequently, we get $$ f_n(x)= \tilde f_n(t) = \frac{(1 + t^2)^n(t^{n + 2} - 1)^2}{t^{2n}(t^2 - 1)^2},\quad\text{if } n\text{ is even}, \\ f_n(x) = \tilde f_n(t) = -\frac{(1 + t^2)^n(1 - t^{n -1})(1 - t^{n+5})}{t^{2n}(t^2 - 1)^2},\quad\text{if } n\text{ is odd}. $$ Hence the only roots of $\tilde f_n$ are the suitable roots of unity and the roots of $f_n$ are all in the real segment $[-4, 0]$ and their closure is the whole segment for large $n$.