<p>If <span>\(a, b, c\)</span> are nonzero real numbers and <span>\(az^2 + bz + c + i = 0\)</span> has purely imaginary roots, then prove that <span>\(a = b^2c\)</span>.</p>
Step-by-Step Solution
Key Concept: If a quadratic with real coefficients has purely imaginary roots, then the roots are of the form ±ki where k is real. Use Vieta's formulas on the roots and the constraint that coefficients are real to establish the relationship between a, b, and c.
<p><strong>Step 1:</strong> Let the purely imaginary roots be z₁ = ki and z₂ = -ki, where k ≠ 0 is real (complex conjugate pair since coefficients are real).</p><p><strong>Step 2:</strong> By Vieta's formulas: Sum of roots: ki + (-ki) = -b/a, so 0 = -b/a, thus <strong>b = 0</strong>.</p><p><strong>Step 3:</strong> Product of roots: (ki)(-ki) = c/a, so -k²i² = c/a, thus k² = c/a (since i² = -1).</p><p><strong>Step 4:</strong> Since z = ki is a root: a(ki)² + b(ki) + c + i = 0. Substituting: a(-k²) + bki + c + i = 0, giving -ak² + c + i(bk + 1) = 0.</p><p><strong>Step 5:</strong> For this complex equation to equal zero with real coefficients: Real part: -ak² + c = 0, so c = ak². Imaginary part: bk + 1 = 0, so k = -1/b.</p><p><strong>Step 6:</strong> Substitute k = -1/b into c = ak²: c = a(-1/b)² = a/b². Therefore: <strong>a = b²c</strong>.</p><p>∴ Proved</p>
Correct Answer: Proof-based