Quadratic Equations
Location of roots
Grade 11
Question:
<p>76. If roots of \(x^2 - (a-3)x + a = 0\) are such that at least one of them is greater than 2, then</p>
<p>(1) \(a \in [7, 9]\)</p>
<p>(2) \(a \in [7, \infty)\)</p>
<p>(3) \(a \in [9, \infty)\)</p>
<p>(4) \(a \in [7, 9)\)</p>
Step-by-Step Solution
Key Concept: For at least one root to exceed 2, we need either f(2) < 0 (roots straddle 2) or f(2) ≥ 0 with vertex beyond 2. The critical condition is f(2) ≤ 0, which gives us the range of a.
<p><strong>Step 1:</strong> Let f(x) = x² - (a-3)x + a. For at least one root > 2, we analyze f(2).</p><p><strong>Step 2:</strong> Calculate f(2) = 4 - 2(a-3) + a = 4 - 2a + 6 + a = 10 - a</p><p><strong>Step 3:</strong> If at least one root > 2, then either:<br/>(i) Both roots > 2, OR<br/>(ii) One root < 2 and one root > 2</p><p><strong>Step 4:</strong> Case (ii): Roots straddle 2 ⟹ f(2) < 0 ⟹ 10 - a < 0 ⟹ a > 10</p><p><strong>Step 5:</strong> Case (i): Both roots > 2 requires f(2) > 0, vertex > 2, and Δ ≥ 0<br/>• f(2) > 0 ⟹ a < 10<br/>• Vertex = (a-3)/2 > 2 ⟹ a > 7<br/>• Δ = (a-3)² - 4a = a² - 10a + 9 = (a-1)(a-9) ≥ 0 ⟹ a ≤ 1 or a ≥ 9</p><p><strong>Step 6:</strong> For case (i): 7 < a < 10 AND (a ≤ 1 or a ≥ 9) ⟹ 9 ≤ a < 10</p><p><strong>Step 7:</strong> Combining both cases: a > 10 OR 9 ≤ a < 10 ⟹ <strong>a ≥ 9</strong></p><p>∴ Answer: a ≥ 9 (or the condition is satisfied when a ∈ [9, ∞))</p>
Correct Answer: 3