Question

In the quadratic equation p(x)=0 with real coefficients has purely imaginary roots. Then, the equation p[p(x)]=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

If a quadratic equation has purely imaginary roots, then the coefficient of x must be equal to zero.
Let $p\left(x\right)=a{x}^2+b$ with a,b of same sign and a,b $ϵ$ R.
Then, $p\left[p\right(x\left)\right]=a(a{x}^2+b{)}^2+b$
p(x) has imaginary roots say ix.
Then, $a{x}^2+bϵR$ and $(a{x}^2+b{)}^2>0$
$\therefore a(a{x}^2+b{)}^2+b\ne 0,\forall xϵR$
Thus, $p\left[p\right(x\left)\right]\ne 0,\forall xϵR$.

Hence, p[p(x)]=0 has complex roots.