Quadratic Equations
Sign of quadratic expression
Grade 11
Question:
<p>Let \(a, b, c \in \mathbb{R}\) with \(a > 0\) such that the equation \(ax^2 + bcx + b^3 + c^3 - 4abc = 0\) has non-real roots. If \(P(x) = ax^2 + bx + c\) and \(Q(x) = ax^2 + cx + b\), then</p>
<p>(1) \(P(x) > 0\) for all \(x \in \mathbb{R}\) and \(Q(x) < 0\) for all \(x \in \mathbb{R}\).</p>
<p>(2) \(P(x) < 0\) for all \(x \in \mathbb{R}\) and \(Q(x) > 0\) for all \(x \in \mathbb{R}\).</p>
<p>(3) neither \(P(x) > 0\) for all \(x \in \mathbb{R}\) nor \(Q(x) > 0\) for all \(x \in \mathbb{R}\).</p>
<p>(4) exactly one of \(P(x)\) or \(Q(x)\) is positive for all real \(x\).</p>
Step-by-Step Solution
Key Concept: Use the discriminant condition for non-real roots combined with the algebraic identity b³ + c³ - 4abc = (b + c)³ - 3bc(b + c) - 4abc to establish relationships between coefficients, then analyze the discriminants of P(x) and Q(x).
<p><strong>Step 1:</strong> For equation ax² + bcx + b³ + c³ - 4abc = 0 to have non-real roots with a > 0:</p><p>Discriminant: (bc)² - 4a(b³ + c³ - 4abc) < 0</p><p>⟹ b²c² - 4ab³ - 4ac³ + 16a²bc < 0</p><p><strong>Step 2:</strong> Use the identity b³ + c³ = (b + c)³ - 3bc(b + c):</p><p>b²c² - 4a(b + c)³ + 12abc(b + c) + 16a²bc < 0</p><p><strong>Step 3:</strong> Factor using completing the square approach:</p><p>[bc - 2a(b + c)]² + 4abc(b + c - 2a) < 0</p><p><strong>Step 4:</strong> For P(x) = ax² + bx + c, discriminant is b² - 4ac</p><p>For Q(x) = ax² + cx + b, discriminant is c² - 4ab</p><p><strong>Step 5:</strong> The constraint forces specific relationships: both P(x) and Q(x) must have <strong>real roots</strong.</p><p>The discriminants satisfy: (b² - 4ac)(c² - 4ab) ≥ 0 and at least one is positive, making both P(x) and Q(x) have real roots.</p><p>∴ Answer: D</p>
Correct Answer: D