Question

The quadratic equation $p\left(x\right)=0$ with real coefficients has purely imaginary roots. Then the equation
$p\left(p\right(x\left)\right)=0$
has

• 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 1

The correct option is D neither real nor purely imaginary roots

Let $P\left(x\right)=a{x}^2+b,a,b\ne 0$ and both have same sign.

$\therefore P\left(x\right)\ne 0\forall x\in R$
Now, $P\left(x\right)$ can take any value real or imaginary but not equal to its roots.
Case $:1$
$P\left(P\right(x\left)\right)=P\left(\text{real value}\right)\ne 0$

Case $:2$
$P\left(P\right(x\left)\right)=P\left(\text{imaginary value but not root of}P\left(x\right)\right)\ne 0$

$\therefore P\left(P\right(x\left)\right)=0$ is not possible for any $x$