<p>If \(a, b, c \in \mathbb{R}\) and \(abc < 0\), then the equation \(bcx^2 + 2(b+c-a)x + a = 0\) has</p>
<p>both positive roots</p>
<p>both negative roots</p>
<p>real roots</p>
<p>one positive and one negative root</p>
Step-by-Step Solution
Key Concept: The condition abc < 0 means exactly one or all three of {a,b,c} are negative. Combined with the discriminant condition, we can determine the nature of roots by analyzing the sign of the quadratic at specific points and the parabola's orientation.
<p><strong>Step 1:</strong> Since abc < 0, either exactly one of {a,b,c} is negative, or all three are negative.</p><p><strong>Step 2:</strong> Evaluate the quadratic at x = 0: f(0) = c. Since abc < 0 and we need to determine root positions, the sign of c matters.</p><p><strong>Step 3:</strong> For a quadratic ax² + bx + c with roots α, β:</p><ul><li>If a > 0 and c > 0: both roots have the same sign (both positive if b < 0, both negative if b > 0)</li><li>If a and c have opposite signs: roots have opposite signs (one positive, one negative)</li></ul><p><strong>Step 4:</strong> Given abc < 0, analyze cases:</p><ul><li><strong>Case 1:</strong> a > 0, b > 0, c < 0 → f(0) = c < 0 and f(∞) → +∞, so roots straddle zero (one positive, one negative) ✓</li><li><strong>Case 2:</strong> a > 0, b < 0, c < 0 → Similar analysis shows roots exist with one positive, one negative ✓</li><li><strong>Case 3:</strong> a < 0, ... → Parabola opens downward; combined with abc < 0 constraints, roots are real ✓</li></ul><p><strong>Step 5:</strong> The discriminant Δ = b² - 4ac: Since a and c have opposite signs (from abc < 0), we have -4ac > 0, so Δ = b² - 4ac > 0. Roots are always real and distinct.</p><p>∴ Answer: C, D (The equation has real roots, and roots have opposite signs or lie in specific intervals depending on the case analysis)</p>
Correct Answer: C,D