Quadratic Equations
Nature of Roots
Grade 11
Question:
<p>Let \(f(x) = x^2 + ax + b; a, b \in \mathbb{R}\). If \(f(1) + f(2) + f(3) = 0\), then the roots of the equation \(f(x) = 0\)</p>
<p>(a) are imaginary</p>
<p>(b) are real and equal</p>
<p>(c) are from the set {1, 2, 3}</p>
<p>(d) real and distinct</p>
Step-by-Step Solution
Key Concept: Use the constraint f(1) + f(2) + f(3) = 0 to find a relationship between coefficients a and b, then analyze the discriminant to determine the nature of roots.
<p><strong>Step 1:</strong> Calculate f(1), f(2), and f(3).</p><p>f(1) = 1 + a + b</p><p>f(2) = 4 + 2a + b</p><p>f(3) = 9 + 3a + b</p><p><strong>Step 2:</strong> Use the constraint f(1) + f(2) + f(3) = 0.</p><p>(1 + a + b) + (4 + 2a + b) + (9 + 3a + b) = 0</p><p>14 + 6a + 3b = 0</p><p>3b = -14 - 6a</p><p>b = -14/3 - 2a</p><p><strong>Step 3:</strong> Analyze the discriminant Δ = a² - 4b.</p><p>Δ = a² - 4(-14/3 - 2a)</p><p>Δ = a² + 56/3 + 8a</p><p>Δ = a² + 8a + 56/3</p><p><strong>Step 4:</strong> Determine the sign of Δ.</p><p>Complete the square: Δ = (a + 4)² - 16 + 56/3</p><p>Δ = (a + 4)² + 56/3 - 16</p><p>Δ = (a + 4)² + 56/3 - 48/3</p><p>Δ = (a + 4)² + 8/3</p><p><strong>Step 5:</strong> Conclude about the roots.</p><p>Since (a + 4)² ≥ 0 for all real a, we have Δ ≥ 8/3 > 0 for all real a.</p><p>When Δ > 0, the quadratic equation f(x) = 0 has two distinct real roots.</p><p><strong>∴ Answer:</strong> d</p>
Correct Answer: d