Quadratic Equations
Symmetric functions of roots
Grade 11
Question:
<p>For the equation \(x^2 - 4\sqrt{2}kx + 2k^4 - 1 = 0\), if \(\alpha + \beta = 4\sqrt{2}k\) and \(\alpha^3 + \beta^3 = 280\sqrt{2}\), then the value of \(k\) is:</p>
<p>\(k = 1\)</p>
<p>\(k = 2\)</p>
<p>\(k = 3\)</p>
<p>\(k = 4\)</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to express α + β and αβ, then apply the identity α³ + β³ = (α + β)³ - 3αβ(α + β) to create an equation in k. The cubic sum formula bypasses finding individual roots.
<p><strong>Step 1:</strong> Apply Vieta's formulas to x² - 4√2kx + 2k⁴ - 1 = 0</p><p>α + β = 4√2k (given)</p><p>αβ = 2k⁴ - 1</p><p><strong>Step 2:</strong> Use the identity α³ + β³ = (α + β)³ - 3αβ(α + β)</p><p>280√2 = (4√2k)³ - 3(2k⁴ - 1)(4√2k)</p><p><strong>Step 3:</strong> Expand (4√2k)³ = 64 · 2√2 · k³ = 128√2k³</p><p>280√2 = 128√2k³ - 12√2k(2k⁴ - 1)</p><p><strong>Step 4:</strong> Divide by √2</p><p>280 = 128k³ - 12k(2k⁴ - 1)</p><p>280 = 128k³ - 24k⁵ + 12k</p><p>24k⁵ - 128k³ - 12k + 280 = 0</p><p><strong>Step 5:</strong> Simplify by dividing by 4</p><p>6k⁵ - 32k³ - 3k + 70 = 0</p><p><strong>Step 6:</strong> Test k = √2</p><p>6(√2)⁵ - 32(√2)³ - 3√2 + 70 = 6(4√2) - 32(2√2) - 3√2 + 70</p><p>= 24√2 - 64√2 - 3√2 + 70 = -43√2 + 70 ≠ 0</p><p>Test k = 2: 6(32) - 32(8) - 6 + 70 = 192 - 256 - 6 + 70 = 0 ✓</p><p>∴ Answer: <strong>k = 2</strong></p>
Correct Answer: B