The quadratic equation $p\left( x \right) = 0$ with real coefficients has purely imaginary roots. Then the equation $p\left( {p\left( x \right)} \right) = 0$ has
A) Only purely imaginary roots
B) All real roots
C) Two real and two imaginary roots
D) Neither real nor purely imaginary roots

Hint: Roots of an equation of variable $x$, is the value of $x$ which satisfies the equation.
The standard form of the quadratic equation is $a{x^2} + bx + c = 0$.
Find the nature of the roots of a quadratic equation by the value of the discriminant, D of the quadratic equation.
Imaginary value (or number) is the value that combines with an iota (denoted by $i$, and equal to $\sqrt { - 1} $ ).
The discriminant of the quadratic equation $a{x^2} + bx + c = 0$ would be ${b^2} - 4ac$ here, a,b and c are coefficients of equation.
Purely imaginary roots indicate that the roots of the equation would only have an imaginary value.
Imaginary roots always occur in conjugate pairs e.g. (a+ib) and (a-ib).

Complete step-by-step answer:
Step 1: Assumption of the roots:
Given that the roots of the given quadratic equation is imaginary.
Let the root of the given equation $p\left( x \right) = 0$ is .
As imaginary roots occur in pairs.
Here, $i$ is iota and $a$ is any real number.
Step 2: Express the equation using roots.
For the roots, $\alpha {\text{ and }}\beta $, the quadratic equation in a variable $x$ is given by
$\left( {x - \alpha } \right)\left( {x - \beta } \right) = 0$
Therefore, For the roots, $ia{\text{ and}} - ia$, the quadratic equation
$\left( {x - ia} \right)\left( {x + ia} \right) = 0$
Hence, $p\left( x \right) = \left( {x - ia} \right)\left( {x + ia} \right) = 0$
Step 3: Obtain the equation $p\left( {p\left( x \right)} \right) = 0$
In the quadratic equation $p\left( x \right) = 0$ , replace all variables $x$ with $p\left( x \right)$ .
$p\left( x \right) = (x - ia)(x + ia) = 0$
For $x = p\left( x \right)$
Equation becomes $p\left( {p\left( x \right)} \right) = \left( {\left( {x - ia} \right)\left( {x + ia} \right) - ia} \right)\left( {\left( {x - ia} \right)\left( {x + ia} \right) + ia} \right) = 0$
Step 4: Simplify the expression and find the root of the equation.
We have, $\left( {\left( {x - ia} \right)\left( {x + ia} \right) - ia} \right)\left( {\left( {x - ia} \right)\left( {x + ia} \right) + ia} \right) = 0$
If the product of two numbers is zero, then either the first number or second number and/or both is zero.
Therefore, $\left( {\left( {x - ia} \right)\left( {x + ia} \right) - ia} \right) = 0$
 And $\left( {\left( {x - ia} \right)\left( {x + ia} \right) + ia} \right) = 0$
The values of x give the solution of the equation. Thus, find out the values of x from both the equations.
$\left( {\left( {x - ia} \right)\left( {x + ia} \right) - ia} \right) = 0$
Expand the multiplication $\left( {x - ia} \right)\left( {x + ia} \right)$
\[
   \Rightarrow \left( {x\left( {x + ia} \right) - ia\left( {x + ia} \right) - ia} \right) = 0 \\
   \Rightarrow {x^2} + iax - iax - {i^2}{a^2} - ia = 0 \\
 \]
Like terms of imaginary numbers can be added and subtracted mathematically. Hence, \[ + iax - iax = 0\].
\[ \Rightarrow {x^2} - {i^2}{a^2} - ia = 0\]
We know, ${i^2} = - 1$
\[
   \Rightarrow {x^2} + {a^2} - ia = 0 \\
   \Rightarrow {x^2} = - {a^2} + ia \\
 \]
$\because x = \pm \sqrt { - {a^2} + ia} $
“x” has a real part as well as an imaginary part. Thus, it is a complex number.
And
$\left( {\left( {x - ia} \right)\left( {x + ia} \right) + ia} \right) = 0$
Expand the multiplication $\left( {x - ia} \right)\left( {x + ia} \right)$
\[
   \Rightarrow \left( {x\left( {x + ia} \right) - ia\left( {x + ia} \right) + ia} \right) = 0 \\
   \Rightarrow {x^2} + iax - iax - {i^2}{a^2} + ia = 0 \\
 \]
Like terms of imaginary numbers can be added and subtracted mathematically. Hence, \[ + iax - iax = 0\].
\[ \Rightarrow {x^2} - {i^2}{a^2} + ia = 0\]
We know, ${i^2} = - 1$
\[
   \Rightarrow {x^2} + {a^2} + ia = 0 \\
   \Rightarrow {x^2} = - {a^2} - ia \\
 \]
$\because x = \pm \sqrt { - {a^2} - ia} $
“x” has a real part as well as an imaginary part. Thus, it is a complex number.
Hence, The values of x obtained here for the equation $\left( {\left( {x - ia} \right)\left( {x + ia} \right) - ia} \right)\left( {\left( {x - ia} \right)\left( {x + ia} \right) + ia} \right) = 0$
are the complex numbers.


The value of $x$ in equation $p\left( {p\left( x \right)} \right) = 0$ neither purely real nor purely imaginary. Thus, option (D) is correct.

Note: The nature of the roots of a quadratic equation is determined by the value of the discriminant (denoted by D).
For equation $a{x^2} + bx + c = 0$
$D = {b^2} - 4ac$ (a, b and c are coefficients of the quadratic equation)
If $D < 0$, Equation would have complex roots (real part+ imaginary part)
If $D > 0$, Equation would have two distinct real roots
If $D = 0$, Equation would have two real and equal roots.
Do not take $\left( {a + ib} \right)$ for purely imaginary root.
A number of the form $\left( {a + ib} \right)$ , where $a{\text{ and }}b$ are real numbers, is called a complex number, $a$ is called the real part, and $b$called the imaginary part.
For the purely real part of the number, the imaginary part is zero.
Obtain roots of the equations by equating expression (or polynomial) to 0.
If the equation $f\left( x \right) = 0$ has 2 roots then the equation $f\left( {f\left( x \right)} \right) = 0$ would have 4 roots.