<p>The quadratic equation \(p(x) = 0\) with real coefficients has purely imaginary roots. Then the equation \(p(p(x)) = 0\) has</p>
<p>only purely imaginary roots</p>
<p>all real roots</p>
<p>two real and two purely imaginary roots</p>
<p>neither real nor purely imaginary roots</p>
Step-by-Step Solution
Key Concept: If p(x) has purely imaginary roots ±bi, then p(x) = a(x² + b²) for some real a > 0. When composing p(p(x)) = 0, we get p(x) = ±bi, leading to real solutions since x² + b² = -b² has real solutions.
<p><strong>Step 1:</strong> Since p(x) = 0 has purely imaginary roots, let them be ±bi where b ∈ ℝ, b ≠ 0.</p><p><strong>Step 2:</strong> Then p(x) = a(x² + b²) where a is a non-zero real coefficient.</p><p><strong>Step 3:</strong> For p(p(x)) = 0, we need p(x) = ±bi.</p><p><strong>Step 4:</strong> This means a(x² + b²) = ±bi, so x² + b² = ±bi/a.</p><p><strong>Step 5:</strong> Rearranging: x² = ±bi/a - b² = -b² ± bi/a, which is a complex number (non-real in general).</p><p><strong>Step 6:</strong> However, solving p(x) = bi: x² = -b² + bi/a has two complex conjugate solutions. Solving p(x) = -bi: x² = -b² - bi/a has two more complex conjugate solutions.</p><p><strong>Step 7:</strong> Therefore, p(p(x)) = 0 has <strong>4 complex (non-real) roots</strong> (all roots are complex since the equation reduces to finding x where x² equals a complex number).</p><p>∴ Answer: D</p>
Correct Answer: D