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Question:

The quadratic equation p(x) = 0 with real coefficients has purely imaginary roots. Then the equation p(p(x)) = 0 has:

All real roots

Neither real nor purely imaginary roots

Only purely imaginary roots

Two real and two purely imaginary roots

Solution:

If the roots are purely imaginary then they are of the form ±iβ. Hence, p(x) = ax² + b with a, b of same sign. p(p(x)) = a(ax² + b)² + b. If x ∈ R or x is purely imaginary ⇒ x² ∈ R ⇒ p(x) ∈ R ⇒ p(p(x)) ≠ 0. Hence, real or purely imaginary number can not satisfy p(p(x)) = 0.