Quadratic Equation and Composite Function

Question

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 \ . $

Answer 1

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?

Answer 2

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!]