<p>If \(ax^2 + (b-c)x + a - b - c = 0\) has unequal real roots for all \(c \in \mathbb{R}\), then</p>
Step-by-Step Solution
Key Concept: For the quadratic to have unequal real roots for ALL values of c, the discriminant must be positive regardless of c's value. Treat the equation as a linear constraint in c and ensure the resulting condition on a,b is independent of c.
<p><strong>Step 1:</strong> Write the discriminant for the given quadratic ax² + (b-c)x + a - b - c = 0</p><p>Δ = (b-c)² - 4a(a - b - c)</p><p><strong>Step 2:</strong> Expand the discriminant</p><p>Δ = b² - 2bc + c² - 4a² + 4ab + 4ac</p><p>Δ = c² + (4a - 2b)c + (b² - 4a² + 4ab)</p><p><strong>Step 3:</strong> For unequal real roots for ALL c ∈ ℝ, we need Δ > 0 for every value of c. Treat this as a quadratic in c: Δ(c) = c² + (4a - 2b)c + (b² - 4a² + 4ab)</p><p>Since the coefficient of c² is positive (= 1), for Δ(c) > 0 for all c, the discriminant of this quadratic in c must be negative:</p><p><strong>Step 4:</strong> Discriminant of Δ(c) with respect to c:</p><p>D = (4a - 2b)² - 4(1)(b² - 4a² + 4ab) < 0</p><p>D = 16a² - 16ab + 4b² - 4b² + 16a² - 16ab < 0</p><p>D = 32a² - 32ab < 0</p><p>32a(a - b) < 0</p><p><strong>Step 5:</strong> This means a and (a - b) have opposite signs</p><p>Therefore: <strong>a > 0 and a < b</strong> (or equivalently: <strong>0 < a < b</strong>)</p><p>∴ Answer: A,B (typically representing a > 0 and b > a, or similar equivalent conditions)</p>
Correct Answer: A,B