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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 4 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 has a common root, then which of the following statement is/are correct.
(A) If b,c ∈ R, then b^2 – 4c ≥ 0.
(B) If P(0) = 1, then p(1) = 0.
(C) equations p(p(p(x))) = 0 and p(p(p(p(p(x))))) = 0 has at least two common root.
(D) zero is root of equation p(p(p(p(p(p(x)))))) = 0

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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?

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  • $\begingroup$ Also by observation since one root is c and p(x) = x^2+bx+c so the product of root will be c . So the other root will be 1 always. For D part if we put zero then the equation at last become p(c)=0. Thank you for helping me out $\endgroup$ Mar 19 at 7:56

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