Quadratic Equations
Roots of Equations
Grade 11
Question:
<p>If \(\alpha\) and \(\beta\) are roots of the equation, \(x^2 - 4\sqrt{2}\,kx + 2e^{4\ln k} - 1 = 0\) for some \(k\), and \(\alpha^2 + \beta^2 = 66\) then \(\alpha^3 + \beta^3\) is equal to</p>
<p>\(248\sqrt{2}\)</p>
<p>\(280\sqrt{2}\)</p>
<p>\(-32\sqrt{2}\)</p>
<p>\(-280\sqrt{2}\)</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to find α + β and αβ, then express α² + β² in terms of these to determine k, and finally compute α³ + β³ using the recurrence relation or algebraic identity.
<p><strong>Step 1: Apply Vieta's formulas</strong></p><p>For x² - 4√2·kx + 2e^(4ln k) - 1 = 0:</p><p>α + β = 4√2·k</p><p>αβ = 2e^(4ln k) - 1 = 2k⁴ - 1</p><p><strong>Step 2: Use the condition α² + β² = 66</strong></p><p>α² + β² = (α + β)² - 2αβ = 66</p><p>(4√2·k)² - 2(2k⁴ - 1) = 66</p><p>32k² - 4k⁴ + 2 = 66</p><p>4k⁴ - 32k² + 64 = 0</p><p>k⁴ - 8k² + 16 = 0</p><p>(k² - 4)² = 0</p><p>k² = 4 ⟹ k = 2 (taking k > 0)</p><p><strong>Step 3: Calculate α + β and αβ</strong></p><p>α + β = 4√2·(2) = 8√2</p><p>αβ = 2(2⁴) - 1 = 2(16) - 1 = 31</p><p><strong>Step 4: Find α³ + β³</strong></p><p>α³ + β³ = (α + β)³ - 3αβ(α + β)</p><p>= (8√2)³ - 3(31)(8√2)</p><p>= 512·2√2 - 744√2</p><p>= 1024√2 - 744√2</p><p>= 280√2</p><p>∴ Answer: A</p>
Correct Answer: A