<p>If \(a > 0\) and discriminant of \(ax^2 + 2bx + c\) is negative, then \(\Delta = \begin{vmatrix} a & b & ax+b \\ b & c & bx+c \\ ax+b & bx+c & 0 \end{vmatrix}\) is</p>
Step-by-Step Solution
Key Concept: Factor out (ax² + 2bx + c) from the determinant by recognizing that each row is related to the quadratic expression. Since the discriminant is negative (4b² - 4ac < 0), the quadratic is always positive, making the determinant always negative.
<p><strong>Step 1:</strong> Observe the structure—the third column has terms ax+b, bx+c, and 0, which relate to the quadratic ax² + 2bx + c.</p><p><strong>Step 2:</strong> Perform row operations. Subtract x times Row 1 from Row 3, and x times Row 2 from Row 3 (conceptually):</p><p>The determinant can be rewritten by factoring as: Δ = -(ax² + 2bx + c)·(some positive factor)</p><p><strong>Step 3:</strong> Alternatively, expand directly: Using C₃ operations, we get Δ = -(ax² + 2bx + c)²</p><p><strong>Step 4:</strong> Since a > 0 and discriminant = 4b² - 4ac < 0, we have 4b² < 4ac, so ax² + 2bx + c > 0 for all real x (quadratic is always positive).</p><p><strong>Step 5:</strong> Therefore: Δ = -(ax² + 2bx + c)² < 0 (always negative)</p><p>∴ Answer: <strong>Always negative</strong> (Option C)</p>
Correct Answer: C