<p>The value of <em>m</em> for which one of the roots of \(x^2 - 3x + 2m = 0\) is double of one of the roots of \(x^2 - x + m = 0\) is</p>
Step-by-Step Solution
Key Concept: If α is a root of the second equation, then 2α is a root of the first equation. Substitute both conditions simultaneously using Vieta's formulas or direct substitution to create a solvable system.
<p><strong>Step 1:</strong> Let α be a root of x² - x + m = 0. Then: α² - α + m = 0 ... (i)</p><p><strong>Step 2:</strong> Given that 2α is a root of x² - 3x + 2m = 0. Then: (2α)² - 3(2α) + 2m = 0, which gives: 4α² - 6α + 2m = 0 ... (ii)</p><p><strong>Step 3:</strong> From equation (i): α² = α - m. Substitute into equation (ii):</p><p>4(α - m) - 6α + 2m = 0</p><p>4α - 4m - 6α + 2m = 0</p><p>-2α - 2m = 0</p><p>α = -m ... (iii)</p><p><strong>Step 4:</strong> Substitute α = -m back into equation (i):</p><p>(-m)² - (-m) + m = 0</p><p>m² + m + m = 0</p><p>m² + 2m = 0</p><p>m(m + 2) = 0</p><p>m = 0 or m = -2</p><p><strong>Step 5:</strong> Check m = 0: Both equations become x² - 3x = 0 and x² - x = 0, sharing root x = 0 (and 2(0) = 0 works). Check m = -2: Equations are x² - 3x - 4 = 0 (roots 4, -1) and x² - x - 2 = 0 (roots 2, -1). Here 2α = 2(2) = 4 ✓</p><p>∴ Answer: <strong>D</strong> (m = -2 or the non-trivial solution)</p>
Correct Answer: D