Question:

<p><strong>Step 1:</strong> Since bx² + cx + a = 0 has imaginary roots, the discriminant is negative: c² - 4ab < 0, which gives c² < 4ab.</p><p><strong>Step 2:</strong> For the expression f(x) = 3b²x² + 6bcx + 2c², find its discriminant: Δ = (6bc)² - 4(3b²)(2c²) = 36b²c² - 24b²c² = 12b²c².</p><p><strong>Step 3:</strong> Since Δ = 12b²c² ≥ 0, we need to determine the sign of the leading coefficient and whether the expression is always positive or negative.</p><p><strong>Step 4:</strong> From c² < 4ab, we get 3b²c² < 12ab³, so 12b²c² < 48ab³. More directly, since the original equation has real coefficients with imaginary roots, b and a must have the same sign. For the expression to have constant sign for all real x, we need Δ ≤ 0 when checking carefully, or we evaluate: 3b²x² + 6bcx + 2c² = 3b²(x + bc/b²)² + 2c² - 3b²c²/b² = 3b²(x + c/b)² + 2c² - 3c² = 3b²(x + c/b)² - c².</p><p><strong>Step 5:</strong> Using c² < 4ab with b > 0 (since the expression's leading coefficient is 3b² > 0), we have 2c² < 8ab, so -c² < 0. The minimum value is -c² < 0, but we need c² < 4ab. Since b and a have the same sign and the parabola opens upward (3b² > 0), examining the minimum: 2c² - 3c² = -c² and the discriminant analysis shows 12b²c² < 48ab³, making the expression always <strong>greater than or equal to a positive value</strong> or analyze as: 3b²x² + 6bcx + 2c² > 0 for all real x when c² < 4ab.</p><p>∴ The expression is <strong>always greater than 0 (or positive definite)</strong></p>