Let $f(x)=x^2+12x+30$. Solve $f(f(f(f(f(x)))))=0$

Here is my solve, is it correct? I figured out that we can restate $f(f(x))$ as $((x+r)(x+s)+r)((x+r)(x+s)+s)$ thus $f(f(f(f(f(x)))))=0$ is $(x+r)^2(x+s)^2(4s+3r)(4r+3s)$

from vieta's

$0=(x+r)(x+s)(36^2-r^2)$ and we get $x=-6$

Did I make a mistake somewhere? I'd also like to see other solutions

EDIT: How can I solve it using vieta then?

Solution 1:

We have $f(x) = (x+6)^2 - 6$, therefore $f^2 (x) = (x+6)^4 - 6$, $f^3 (x) = (x+6)^8 - 6$, and so on.

So $f^5 (x) = (x+6)^{32} - 6 = 0$, giving the (real) solutions $x=-6 \pm \sqrt[32]{6}$.

Solution 2:

Define inductively $f^n(x)=f(f^{n-1}(x))$ for all $n\geq1$, so that you are solving $f^5(x)=0$.

We have $f^5(x)=f(f^4(x))$ thus $f^4(x)$ is either solution of $x^2+12x+30=0$, i.e. $$ f^4(x)=-3\pm\sqrt{6}. $$ Then for the same reason $f^3(x)$ is any value $z$ such that $$ z^2+12z+30=-3\pm\sqrt{6}. $$ Iterating this process a few more times you get to the solutions of $f^5(x)=0$.