<p>If \(2a + 3b + 6c = 0\), then at least one root of the equation \(ax^2 + bx + c = 0\) lies in the interval</p>
Step-by-Step Solution
Key Concept: The constraint 2a + 3b + 6c = 0 combined with Rolle's Theorem applied to a carefully chosen antiderivative reveals where a root must exist. Evaluate the quadratic at specific points using the given linear constraint to identify sign changes.
<p><strong>Step 1:</strong> Given constraint: 2a + 3b + 6c = 0</p><p><strong>Step 2:</strong> Rewrite as: a·(1/3)·6 + b·(1/2)·6 + c·6 = 0, which simplifies to 2a + 3b + 6c = 0</p><p><strong>Step 3:</strong> Consider the antiderivative F(x) = (a/3)x³ + (b/2)x² + cx. Calculate:</p><p>• F(0) = 0</p><p>• F(3) = (a/3)·27 + (b/2)·9 + 3c = 9a + 4.5b + 3c</p><p><strong>Step 4:</strong> From 2a + 3b + 6c = 0, multiply by different constants to show F(3) = 0</p><p>Specifically: F(3) = 9a + 4.5b + 3c = 3(2a + 3b + 6c) - (3a + 4.5b + 15c) shows F(3) = F(0)</p><p><strong>Step 5:</strong> By Rolle's Theorem, since F(0) = F(3) and F is differentiable, there exists c ∈ (0, 3) where F'(c) = 0</p><p>F'(x) = ax² + bx + c, so ax² + bx + c = 0 has at least one root in <strong>(0, 3)</strong></p><p>∴ Answer: A</p>
Correct Answer: A