<s>[INST] if $~f(x)=x^2+10x+20~$, find the number of solutions of $f(f(f(f(x))))=0$ or $f^4(x)=0$ where $f^n$ means f is composed to itself n times. The answer given is 2.  I tried to bash the question by $~f^3(x)=\alpha~$ where $~\alpha~$ is the root. Then find the solution, and keep on continuing. It is a long process, so I ditched the idea. I'm looking for a short and elegant solution.
 [/INST] To do it the very hard way
$a^2 + 10a + 20 = k$ was solutions $a =\frac{-10\pm\sqrt{100-4(20-k)}}2=-5\pm \frac {\sqrt{20 + 4k}}2=-5\pm \sqrt{5+k}$.
So if $f^4(f^3(x)) = 0$ we have solutions $f^3(x)=-5\pm \sqrt{5}$.
To solve $f^3(x) =f(f^2(x))=-5\pm \sqrt{5}$ we have solutions $f^2(x) = -5\pm\sqrt{5 + (-5\pm\sqrt 5)}=-5\pm \sqrt{\pm \sqrt 5} =-5\pm \sqrt{\sqrt 5}=-5\pm \sqrt[4] 5$.
To solve $f^2(x) =f(f(x)) = -5\pm \sqrt[4] 5$ we have solutions $f(x) = -5\pm\sqrt{5 -(-5\pm \sqrt[4]5)} = -5\pm \sqrt[8] 5$ (definitely seeing a pattern here).
And finally to solve $f(x) = -5\pm \sqrt[8]5$ we have solutions $x = -5\pm \sqrt{5+(-5\pm \sqrt[8]5)} =-5\pm \sqrt[16]5$.  Two solutions.
(And by induction the solution solution to $f^n(x)=0$ is $x = -5 \pm \sqrt[2^n]5$)
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