The quadratic equation `p(x)=0` with real coefficients has purely imaginary roots. Then the equation `p(p(x))=0` has only purely imaginary roots at real roots two real and purely imaginary roots neither real nor purely imaginary roots

Options

A. only purely imaginary roots
B. all real roots
C. two real and two purely imaginary roots
D. neither real nor purely imaginary roots

Answer

D

Explanation

Let `p(x) = ax^(2) + bx + c, "where" a, b, c in R`. It is given that p(x) = 0 has purely imaginary roots. Therefore, b=0 and a and c are non-zero real numbers of the same sign.
Putting `b = 0 "in" p(x) = ax^(2) + bx + x`, we obtain `p(x) = ax^(2) + c`.
`therefore" "p(p(x)) = p(ax^(2) + c) = a(ax^(2) + c)^(2) + c ne 0`
If `x in R`, then `p(p(x)) = a(ax^(2)+c)^(2) + c ne 0` If x is purely imaginary, say i `alpha`, where `a in R`. Then,
`p(p(i alpha))=a(-a alpha^(2)+c)^(2)+c ne 0" "[{:(therefore" ""a and c are of"),(" ""the same sign"):}]`
Thus, p(p(x)) = 0 has neither real non purely imaginary roots