Quadratic Equations
Roots and intervals
Grade 11
Question:
<p>If the roots of the equation \(x^2 + 2ax + b = 0\) are real and distinct and they differ by at most \(2m\), then \(b\) lies in the interval</p>
<p>\((a^2, a^2 + m^2)\)</p>
<p>\((a^2 - m^2, a^2)\)</p>
<p>\([a^2 - m^2, a^2)\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Use the discriminant condition for real distinct roots and the constraint that |r₁ - r₂| ≤ 2m to establish bounds on b. The key is expressing the root difference as √(discriminant)/|1| and relating it to the parameter m.
<p><strong>Step 1:</strong> For real and distinct roots: Δ = 4a² - 4b > 0, so <strong>a² > b</strong></p><p><strong>Step 2:</strong> The roots are r₁, r₂. We know |r₁ - r₂| = √(Δ)/|coefficient of x²| = √(4a² - 4b)/1 = 2√(a² - b)</p><p><strong>Step 3:</strong> Given constraint: |r₁ - r₂| ≤ 2m means 2√(a² - b) ≤ 2m</p><p><strong>Step 4:</strong> Squaring both sides: √(a² - b) ≤ m → a² - b ≤ m²</p><p><strong>Step 5:</strong> Rearranging: <strong>b ≥ a² - m²</strong></p><p><strong>Step 6:</strong> Combining conditions from Step 1 and Step 5: <strong>a² - m² ≤ b < a²</strong></p><p>∴ Answer: C</p>
Correct Answer: C