Quadratic Equations
Nature of roots
Grade 11
Question:
<p>Given that the equation \(x^2 - ax + 3 - b = 0\) has two distinct real roots, \(x^2 + (6-a)x + 6 - b = 0\) has two equal real roots, and \(x^2 + (4-a)x + 5 - b = 0\) has no real roots. Then the ranges of a and b are</p>
<p>(a) \(2 < a < 4,\ 2 < b < 5\)</p>
<p>(b) \(1 < a < 4,\ 2 < b < 5\)</p>
<p>(c) \(2 < a < 4,\ 1 < b < 5\)</p>
<p>(d) \(1 < a < 4,\ 1 < b < 5\)</p>
Step-by-Step Solution
Key Concept: Use the discriminant conditions (Δ > 0, Δ = 0, Δ < 0) for each equation to create a system of inequalities, then find the intersection of all three regions simultaneously.
<p><strong>Step 1: Apply discriminant condition to equation 1 (two distinct real roots)</strong></p><p>For x² - ax + 3 - b = 0: Δ₁ > 0</p><p>a² - 4(3 - b) > 0</p><p>a² - 12 + 4b > 0</p><p><strong>Condition 1:</strong> a² + 4b > 12</p><p><strong>Step 2: Apply discriminant condition to equation 2 (two equal real roots)</strong></p><p>For x² + (6-a)x + 6 - b = 0: Δ₂ = 0</p><p>(6-a)² - 4(6 - b) = 0</p><p>36 - 12a + a² - 24 + 4b = 0</p><p>a² - 12a + 4b + 12 = 0</p><p><strong>Condition 2:</strong> a² - 12a + 4b = -12</p><p><strong>Step 3: Apply discriminant condition to equation 3 (no real roots)</strong></p><p>For x² + (4-a)x + 5 - b = 0: Δ₃ < 0</p><p>(4-a)² - 4(5 - b) < 0</p><p>16 - 8a + a² - 20 + 4b < 0</p><p>a² - 8a + 4b - 4 < 0</p><p><strong>Condition 3:</strong> a² - 8a + 4b < 4</p><p><strong>Step 4: Solve the system</strong></p><p>From Condition 2: 4b = -12 - a² + 12a</p><p>Substitute into Condition 1:</p><p>a² + (-12 - a² + 12a) > 12</p><p>12a - 12 > 12</p><p>12a > 24</p><p>a > 2</p><p>Substitute into Condition 3:</p><p>a² - 8a + (-12 - a² + 12a) < 4</p><p>4a - 12 < 4</p><p>4a < 16</p><p>a < 4</p><p>From Condition 2: b = (-a² + 12a - 12)/4 = (12a - a² - 12)/4</p><p><strong>∴ Answer A: 2 < a < 4 and b = (12a - a² - 12)/4 (or equivalent form)</strong></p>
Correct Answer: A