Quadratic Equation and Composite Function I am trying to solve this but I think enough information is not given for example in (A): it will only hold true if we do not take $ \ x \ $ as imaginary .
(All four are given as correct in answer) .
So can anyone help me with this question?
Let quadratic equation $ \ p(x) \ = \ 0 \ $ (where $ \ p(x) \ = \ x^2 + bx + c \ ) \ $ and equation $ \ p(p(p(x))) \ = \ 0 \ $ have a common
root, then which of the following statements is/are correct:
(A) if $ \ b,c \ \in \mathbb{ R}, \ $ then $ \  b^2 – 4c \ \ge \  0 \ .   $
(B) if $ \ p(0) \ = \  1 \ , \ $ then $ \ p(1) \ = \  0 \ .  $
(C) the equations $ \ p(p(p(x))) \ = \  0 \ $ and $ \  p(p(p(p(p(x))))) \ = \  \ $0 have at least two common roots.
(D) zero is a root of the equation $ \ p(p(p(p(p(p(x)))))) \ = \ 0 \ . $
 A: Suppose $x_0$ is the common root of the equation $p(x)=0$ and $p(p(p(x)))=0$. Then we have $p(p(p(x_0)))=p(p(0))=p(c)=0$. This means that $c$ is a root of $p(x)=0$.
A) If $b,c \in \mathbb{R}$, then the quadratic will have a real root (namely, $c$), hence it will have both roots real, hence the discriminant is positive.
B) This is just if $c=1$. We know $p(c)=0$.
C) You can see that $x_0$ and $c$ are both common roots here.
D) Can you see why this is true now?
A: One way to approach this sort of problem about repeated function composition is to consider the "mapping" properties of the function.  As a quadratic polynomial, we know that $ \ p(x) \ $  will "map" its two "zeroes" $ \ r \ , \ s \ $ (possibly both the same real number, possibly two complex numbers) to zero [   $ \ r,s \   \xrightarrow{p} \ 0 \ ] \ $ and zero to its "$ \ y-$intercept" $ \ c \ \ [ 0 \   \xrightarrow{p} \ c \ ] \ \ . $
We are given the information about applying this function as an iterated map that $ \ p(p(p(x))) \ = \ 0 \ \ . \ $  Considered "schematically", if we start with one of the zeroes, this equation describes a sequence
$$ r,s \ \  \xrightarrow{p} \ \ 0 \ \   \xrightarrow{p} \ \  c \ \   \xrightarrow{p} \ \  0 \ \ . \ $$  But since $ \ p(x) \ $ only has two zeroes, this indicates that $ \ c \ $ must be one of them and that the numbers in the set $ \ \{ \ 0 \ , \ c  \ \}  \ $ are involved in a "cycle" of just two steps.  (This makes, say,  $ \ r \ = \ c \ $ the "common root" in the problem statement; it is also true that $ \ p(p(p(s))) \ = \ 0 \ \ , \ $ but we will not be concerned with the rôle of the "other zero" (just yet).)
To judge from what is given in the problem statement, the zeroes of $ \ p(x) \ $ need not be distinct. But if its coefficients are real, which means $ \ c \ $ is real, then the zeroes are not a "complex-conjugate pair".  So the discriminant of $ \ p(x) \ $ has to be non-negative, or $ \ b^2 - 4c \ \ge \ 0 \ \ . \ \mathbf{[A]} $
The cycle of two steps resolves the rest of the choices.  If $ \ p(0) \ = \ 1 \ \ , \ $ it is evident that $ \ c \ $ would equal $ \ 1 \ \ , \ $ so $ \ \mathbf{[B]} \ $ is a correct statement.  For $ \ \mathbf{[C]} \ , \ $ the difference between the first and second equations is an additional two "levels" of function composition, which is equivalent to one more cycle around the mapping, so the two zeroes $ \ c \ , \ s \ $ are the "two common roots".  Finally, the two-step cycle indicates that the six-times-iterated-$ p $ equation has the same root as the four-times-iterated-$ p $ equation, which in turn has the same root as $ \ p(p(x)) \ = \ 0 \ \ , \ $ namely, $ \ 0 \ \ . \ \mathbf{[D]} \ \ . $  So all four statements are correct.
$$ \ \ $$
If we put together what we are given and what we've worked out, we find that $ \ p(x) \ = \ (x - c)·(x - s) \ = \ x^2 \ - \ (c + s)·x \ + \ cs \ \ ; \ $ since $ \ p(0) \ = \ c \ \ , \ $ it must be the case that $ \ s \ = \ 1 \ $ is the "second zero" and  $ \ p(x) \ = \   x^2 \ - \ (c + 1)·x \ + \ c \ \ .   $
With some amount of labor (or computational aid), we obtain
$$ p(p(x)) \ \ = \ \ x^4 \ - \ 2·( c + 1 )·x^3 \ + \ c·(c + 3)·x^2 \ + \ ( 1 - c^2)·x \ \ . \  $$
It is evident that $ \ p(p(0)) \ = \ 0 \ $ and  that
$$ p(p(c)) \ \ = \ \ c^4 \ - \ 2c^4 \ - \ 2c^3 \ + \ c^4 \ + \ 3c^3 \ + \ c \ \ - \ c^3 \ \ = \ \ c \ \ ,    $$
as the two-step cycle requires.  But we also observe that
$$ p(p(1)) \ \ = \ \ 1^4 \ - \ 2c \ - \ 2  \ + \ c^2 \ + \ 3c   \ + \ 1 \ - \ c^2 \ \ = \ \ c \ \ .    $$
Hence, we discover that $ \ c \ = \ r \ = \ s \ = \ 1 \ \ , \ $ and so   $ \ p(x) \ $ has the "double zero" $ \ 1 \ \ , \ $ giving us $ \ p(x) \ = \ (x - 1)^2 \ \   $ (and $ \ p(p(x)) \ = \ x^2·(x - 2)^2 \ \ ) \ . \ $  The function map is thus simply described by $ \ p \ : \ \ 0 \ \longleftrightarrow \ 1 \ \ . $  [Now the truth of the four propositions follows immediately!]
