Let, $x + \frac{1}{x} + 4 = t$

$\therefore x + \frac{1}{x} = t - 4$

$ \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = {t^2} - 8t + 16$

$ \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 14 = {t^2} - 8t = t(t - 8)$

$\therefore f(t) = t(t - 8)$

So, $f(0) = 0$

by IIT Alumnus

$\therefore x + \frac{1}{x} = t - 4$

$ \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = {t^2} - 8t + 16$

$ \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 14 = {t^2} - 8t = t(t - 8)$

$\therefore f(t) = t(t - 8)$

So, $f(0) = 0$