<p>If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2 - x + 1 = 0\), then \(\alpha^{2009} + \beta^{2009} =\)</p>
Step-by-Step Solution
Key Concept: The roots of x² - x + 1 = 0 are complex cube roots of unity (ω and ω²), which satisfy ω³ = 1. Use this periodicity to reduce the exponent 2009 modulo 3.
<p><strong>Step 1:</strong> Find the roots of x² - x + 1 = 0 using the quadratic formula:<br>x = (1 ± √(1-4))/2 = (1 ± i√3)/2</p><p><strong>Step 2:</strong> Recognize these as ω = e^(2πi/3) and ω² = e^(4πi/3), the complex cube roots of unity (excluding 1). These satisfy ω³ = 1 and 1 + ω + ω² = 0.</p><p><strong>Step 3:</strong> Since α³ = 1 and β³ = 1, we need to find 2009 (mod 3):<br>2009 = 3 × 669 + 2, so 2009 ≡ 2 (mod 3)</p><p><strong>Step 4:</strong> Therefore α^2009 = α² and β^2009 = β²<br>α² + β² = (α + β)² - 2αβ</p><p><strong>Step 5:</strong> From Vieta's formulas: α + β = 1 and αβ = 1<br>α² + β² = 1² - 2(1) = -1</p><p>∴ Answer: B (which is -1)</p>
Correct Answer: B