Step-by-Step Solution:

1. Analyze the quadratic polynomial p(x)

   Let the quadratic equation be $p(x) = ax^2 + bx + c = 0$, where the coefficients $a, b, c$ are real numbers and $a \neq 0$.

   The problem states that the roots of $p(x) = 0$ are purely imaginary. Let the roots be $ki$ and $-ki$ for some real number $k \neq 0$. Since the coefficients are real, any non-real roots must occur in conjugate pairs. The conjugate of $ki$ is $-ki$, which is consistent.

   Using Vieta's formulas:

   • Sum of roots: $ki + (-ki) = 0$. Also, the sum of roots is $-b/a$. So, $-b/a = 0$, which implies $b=0$.
   • Product of roots: $(ki)(-ki) = -k^2i^2 = k^2$. Also, the product of roots is $c/a$. So, $c/a = k^2$.

   Since $k$ is a non-zero real number, $k^2 > 0$. Therefore, $c/a > 0$, which means $a$ and $c$ have the same sign.

   With $b=0$, the polynomial is $p(x) = ax^2 + c$. We can write this as $p(x) = a(x^2 + c/a)$. Let $\alpha = c/a = k^2$. Since $k \neq 0$, we have $\alpha > 0$. So, the polynomial is of the form $p(x) = a(x^2 + \alpha)$ where $a \in \mathbb{R}, a \neq 0$ and $\alpha > 0$. The roots of $p(x)=0$ are $x = \pm i\sqrt{\alpha}$.

2. Analyze the equation p(p(x)) = 0

   The equation $p(p(x))=0$ means that the value of $p(x)$ must be a root of the equation $p(y)=0$. Let $y = p(x)$, then we have $p(y)=0$.

   From step 1, we know the roots of $p(y)=0$ are $y = i\sqrt{\alpha}$ and $y = -i\sqrt{\alpha}$.

   Therefore, we must solve for $x$ in the following two equations: (i) $p(x) = i\sqrt{\alpha}$ (ii) $p(x) = -i\sqrt{\alpha}$

3. Solve for the roots of p(p(x)) = 0

   Let's solve each equation for $x$ using the form $p(x) = a(x^2 + \alpha)$.

   For equation (i): $a(x^2 + \alpha) = i\sqrt{\alpha}$ $x^2 + \alpha = \frac{i\sqrt{\alpha}}{a}$ $x^2 = -\alpha + i \frac{\sqrt{\alpha}}{a}$

   For equation (ii): $a(x^2 + \alpha) = -i\sqrt{\alpha}$ $x^2 + \alpha = -\frac{i\sqrt{\alpha}}{a}$ $x^2 = -\alpha - i \frac{\sqrt{\alpha}}{a}$

4. Characterize the roots

   In both cases, $x^2$ is equal to a complex number of the form $R + iI$ where the imaginary part $I$ is non-zero (since $a \neq 0$ and $\alpha > 0$). Let's denote this complex number as $Z$. So we have $x^2=Z$.

   Let's analyze the nature of $x$ when $x^2 = Z$, where $Z$ is a complex number that is not purely real.

   • If $x$ were a real number, then $x^2$ would be a real number. This contradicts the fact that $Z$ has a non-zero imaginary part.
   • If $x$ were a purely imaginary number, say $x=iy$ for some real $y$, then $x^2 = (iy)^2 = -y^2$. This would be a non-positive real number. This also contradicts the fact that $Z$ has a non-zero imaginary part.

   Therefore, the roots $x$ cannot be real and cannot be purely imaginary. The roots $x$ must be complex numbers with both non-zero real and non-zero imaginary parts.

   For instance, let $x=u+iv$. Then $x^2 = (u^2-v^2) + 2uvi$. For $x^2 = -\alpha + i \frac{\sqrt{\alpha}}{a}$, we have $2uv = \frac{\sqrt{\alpha}}{a}$. Since the right side is non-zero, neither $u$ nor $v$ can be zero.

   The equation $p(p(x))=0$ is a quartic equation (a polynomial of degree 4), so it has four roots in the complex numbers. Both quadratic equations $x^2 = -\alpha + i \frac{\sqrt{\alpha}}{a}$ and $x^2 = -\alpha - i \frac{\sqrt{\alpha}}{a}$ will yield two roots each. As we have shown, none of these four roots are real or purely imaginary.

5. Conclusion

   The roots of the equation $p(p(x))=0$ are neither real nor purely imaginary. This matches option D.

Evaluation of Options:

• A: one purely imaginary root - Incorrect. No roots are purely imaginary.
• B: all real roots - Incorrect. No roots are real.
• C: two real and two purely imaginary roots - Incorrect. No roots are real or purely imaginary.
• D: neither real nor purely imaginary roots - Correct.