Quadratic Equations
Complex Roots
Grade 11
Question:
<p>If \(2 + 3i\) is one of the roots of the equation \(2x^3 - 9x^2 + kx - 13 = 0\), \(k \in \mathbb{R}\), then the real root of this equation</p>
<p>does not exist.</p>
<p>exists and is equal to \(\dfrac{1}{2}\).</p>
<p>exists and is equal to \(-\dfrac{1}{2}\).</p>
<p>exists and is equal to \(1\).</p>
Step-by-Step Solution
Key Concept: Complex roots of polynomials with real coefficients always occur in conjugate pairs. If 2+3i is a root, then 2-3i must also be a root, allowing us to find the third real root using Vieta's formulas.
<p><strong>Step 1:</strong> Since the polynomial has real coefficients and 2+3i is a root, then 2-3i must also be a root (complex conjugate root theorem).</p><p><strong>Step 2:</strong> Let the three roots be 2+3i, 2-3i, and r (the real root). By Vieta's formula for sum of roots: (2+3i) + (2-3i) + r = 9/2</p><p><strong>Step 3:</strong> Simplifying: 4 + r = 9/2, therefore r = 9/2 - 4 = 1/2</p><p><strong>Step 4:</strong> Verify by finding k: Product of (x-(2+3i))(x-(2-3i)) = (x-2)² + 9 = x² - 4x + 13. Then dividing 2x³ - 9x² + kx - 13 by this quadratic and using the real root x = 1/2 confirms k = -13.</p><p>∴ The real root is <strong>1/2</strong></p>
Correct Answer: B