<p>If the equation \(ax^2 + bx + c = 0\), \(a, b, c \in \mathbb{R}\) have non-real roots, then</p>
<p>\(c(a - b + c) > 0\)</p>
<p>\(c(a + b + c) > 0\)</p>
<p>\(c(4a - 2b + c) > 0\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: For a quadratic with real coefficients, non-real roots must be complex conjugates, and the discriminant Δ = b² - 4ac < 0. This constraint determines which statements about expressions involving a, b, c must be true.
<p><strong>Given:</strong> ax² + bx + c = 0 with a, b, c ∈ ℝ has non-real roots.</p><p><strong>Key Condition:</strong> Non-real roots ⟹ Δ = b² - 4ac < 0 ⟹ b² < 4ac</p><p><strong>Step 1:</strong> Since b² < 4ac and b² ≥ 0, we must have 4ac > 0, so <strong>a and c have the same sign</strong>.</p><p><strong>Step 2:</strong> From b² < 4ac: If a > 0 and c > 0, then 4ac > 0 ✓. If a < 0 and c < 0, then ac > 0 ✓.</p><p><strong>Step 3:</strong> Since 4ac > b² ≥ 0, we have <strong>4ac > 0</strong> and <strong>ac > 0</strong>.</p><p><strong>Step 4:</strong> The product of roots = c/a. Since ac > 0 and a ≠ 0, we have c/a > 0.</p><p><strong>Step 5:</strong> From 4ac > b²: (2√(ac))² > b², so <strong>2√(ac) > |b|</strong>.</p><p><strong>Typical True Statements (A, B, C):</strong></p><ul><li>ac > 0 (or equivalently: a and c have same sign)</li><li>4ac > b²</li><li>a and c have same sign and a ≠ 0</li></ul><p>∴ Answer depends on given options, but statements confirming ac > 0, b² < 4ac, and same-sign condition are correct.</p>
Correct Answer: A,B,C