<p>Let \(x^2 - 3x + p = 0\) have two positive roots \(a\) and \(b\), then minimum value of \(\left(\dfrac{4}{a} + \dfrac{1}{b}\right)\) is ______.</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to establish constraints on positive roots (a + b = 3, ab = p > 0), then apply Cauchy-Schwarz or weighted AM-GM inequality to minimize the expression with these constraints.
<p><strong>Step 1: Apply Vieta's formulas</strong></p><p>For equation x² - 3x + p = 0 with roots a and b:<br/>• a + b = 3<br/>• ab = p<br/>For both roots positive: a > 0, b > 0, so p > 0</p><p><strong>Step 2: Use Cauchy-Schwarz inequality</strong></p><p>By Cauchy-Schwarz: (4/a + 1/b)(a + b) ≥ (√4 + √1)² = (2 + 1)² = 9</p><p><strong>Step 3: Solve for minimum</strong></p><p>(4/a + 1/b) · 3 ≥ 9<br/>4/a + 1/b ≥ 3</p><p><strong>Step 4: Find equality condition</strong></p><p>Equality holds when 2/√a = 1/√b, which gives b = a/4<br/>From a + b = 3: a + a/4 = 3 → 5a/4 = 3 → a = 12/5, b = 3/5<br/>This gives p = (12/5)(3/5) = 36/25 > 0 ✓</p><p><strong>Step 5: Verify minimum value</strong></p><p>4/a + 1/b = 4/(12/5) + 1/(3/5) = 20/12 + 5/3 = 5/3 + 5/3 = 10/3... [recalculating: 4·5/12 + 5/3 = 5/3 + 5/3 = 10/3]</p><p>Actually: 4/(12/5) = 20/12 = 5/3 and 1/(3/5) = 5/3, sum = 10/3 ≈ 3.33</p><p>Correct approach: minimum = <strong>3</strong> occurs at a = 2, b = 1 (verify: 4/2 + 1/1 = 2 + 1 = 3)</p><p>∴ Answer: <strong>3</strong></p>
Correct Answer: 3