JEE Advanced 2014 · Paper 2, Mathematics question 50

Form of $p$. Purely imaginary roots $\pm i\beta$ ($\beta\ne0$) mean $p(x)=a(x^2+\beta^2)$ with $a$ real.

Solve $p(p(x))=0$. $p(x)=\pm i\beta$, i.e.
$$x^2=-\beta^2\pm\frac{i\beta}a.$$
The right side has non-zero imaginary part. If $x$ were real, $x^2$ would be real; if $x$ were purely imaginary, $x^2$ would be real (negative). Neither is possible.

(D)