<s>[INST] number of solutions of $f(f(f(f(x))))$ Let $f(x)={x^2+10x+20}$ then the number of real solutions of $f(f(f(f(x))))=0$
My try: I first tried finding the roots of $f(x)=0$ they came out to be
$-5+\sqrt{5}$ and $-5-\sqrt{5}$.
Let these roots be $\alpha$ and $\beta$
thus $f(\alpha)=0$ also if $f(f(f(f(x))))=0$ then $f(f(f(x)))=\alpha$
I was able to go only till this. I skipped the method of actually finding the composite function as it would be too cumbersome. Any ideas, helps will be appreciated.
 [/INST] The multiple composition of the equation can also be viewed in this way (amounting to a different interpretation of what is described in many of the other posted answers).  The function $ \ f(x) \ = \ x^2 + 10x + 20 \ = \ (x + 5)^2 - 5 \ \ $ has the two real zeroes $ \ r_{\pm} \ = \ -5 \ \pm \sqrt{5} \ \ , $ so the next level of composition, $ \ f( \ f(x) \ ) \ = \ 0 \ \ $ requires us to solve $ \ f(x) \ = \ -5 \ \pm \ \sqrt{5} \ \ ; $ further levels of composition seem to call for more daunting calculation.
We might instead look at how the function transforms numbers.  The fixed points of the function are given by
$$ f(x) \ = \  x \ \ \Rightarrow \ \ x^2 \ + \ 10x \ + \  20 \ \ = \ \ x \ \ \Rightarrow \ \ x^2 \ + \ 9x \ + \  20 \ \ = \ \ (x + 5)·(x + 4) \ \ = \ \ 0 $$
$$ \Rightarrow \ \ x \ \ = \ \ -4 \ , \ -5 \ \ . $$
So $ \ f(-4) \ = \ -4 \ \ $ and $ \ f(-5) \ = \ -5 \ \ . $  We could then ask about how other numbers are "mapped" by the function relative to the fixed points.  For $ \ x \ = \ -5 + \alpha \ \ , $ we obtain
$$   ( \ [-5 + \alpha] + 5)^2 - 5 \ \ = \ \ \alpha^2 \ - \ 5 \ \ , $$
We see from this that $ \ f(x) \ = \ 0 \ \ $ corresponds to the mapping of $ \ -5 + \alpha \ \rightarrow \ 0 \ \ $ such that $ \ \alpha^2 - 5 \ = \ 0 \ \ . $  Hence there are two such points, $ \ -5 \pm \sqrt5 \ \ , $ which are symmetrically located around the axis of symmetry of the parabola representing the quadratic polynomial (as we expect of the zeroes).  What we now also see is that $ \ f( \ f(x) \ ) \ = \ 0 \ $ corresponds to $ \ -5 + \alpha' \ \rightarrow \ -5 + \sqrt5 \ \ $ such that $ \ (\alpha')^2 - 5 \ = \ ( \ \alpha^2 \ )^2 - 5 \ = \ 0 \ \ . $
Repeated applications of the function transformation maintain the symmetry of the two points about $ \ x = -5 \ \ $ while "compressing" them successively closer to this fixed point (which would be referred to as an "attractor").  As also described in comments and the other posted answers, we observe that $ \ f^3( x ) \ = \ 0 \ $ is solved by  $ \ -5 + \alpha'' \ \rightarrow \ -5 + \alpha' \ \ $ with $ \ (\alpha'')^2 - 5 \ = \ ( \ ( \ \alpha^2 \ )^2 \ )  ^2 - 5 \ = \ 0 \ \  $  and finally $ \ f^4( x ) \ = \ 0 \ $ is solved by  $ \ -5 + \alpha'''  \ \ $ with $ \ (\alpha''')^2 - 5 \ = \ ( \ ( \ ( \ ( \ \alpha^2\ )^2 \ )^2 \ )  ^2 - 5 \ = \ 0 \ \Rightarrow \ \alpha''' \ = \ \pm 5^{1/16} \ \ . $  At any finite number of compositions then, we continue to obtain just two real zeroes symmetrically placed about $ \ -5 \ $ (and a whole slew of complex zeroes).
What about the other fixed point?  If we use $ \ x \ = \ -4 + \beta \ \ , $ we obtain
$$   ( \ [-4 + \beta] + 5)^2 - 5 \ \ = \ \ ( \ 1 + \beta \ )^2 \ - \ 5 \ \ = \ \ 0 \ \ \Rightarrow \ \ \beta \ \ = \ \ -1 \pm \sqrt5 \ \ , $$
leading us again to $ \ x \ = \ -5 \pm   \sqrt5 \ \ . $  (So the fixed point $ \ -4 \ $ acts as a "repellor".)
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