Quadratic Equations
Roots of Quadratic Equation
Grade 11
Question:
<p>Given one root of \(f(x)\) is \(-1\). Then \(f(x) = a(x+1)(x-\alpha)\), where \(\alpha\) is the other root of the quadratic equation. If \(f(1) + f(2) = 0\), find \(\alpha\).</p>
<p>\(\dfrac{8}{5}\)</p>
<p>\(\dfrac{5}{8}\)</p>
<p>\(\dfrac{3}{5}\)</p>
<p>\(\dfrac{5}{3}\)</p>
Step-by-Step Solution
Key Concept: Use the given root to express the quadratic in factored form f(x) = a(x+1)(x-α), then apply the condition f(1) + f(2) = 0 to find the unknown root α.
<p><strong>Step 1:</strong> Express f(x) using the known root. Since -1 is a root, write:</p><p>f(x) = a(x+1)(x-α) where α is the other root and a ≠ 0</p><p><strong>Step 2:</strong> Calculate f(1):</p><p>f(1) = a(1+1)(1-α) = a(2)(1-α) = 2a(1-α)</p><p><strong>Step 3:</strong> Calculate f(2):</p><p>f(2) = a(2+1)(2-α) = a(3)(2-α) = 3a(2-α)</p><p><strong>Step 4:</strong> Apply the condition f(1) + f(2) = 0:</p><p>2a(1-α) + 3a(2-α) = 0</p><p>a[2(1-α) + 3(2-α)] = 0</p><p>Since a ≠ 0: 2(1-α) + 3(2-α) = 0</p><p><strong>Step 5:</strong> Expand and solve for α:</p><p>2 - 2α + 6 - 3α = 0</p><p>8 - 5α = 0</p><p>α = 8/5</p><p>∴ Answer: <strong>α = 8/5</strong></p>
Correct Answer: A