<p>Find the number of pairs (a, b) of real numbers such that whenever \(\alpha\) is a root of \(x^2 + ax + b = 0\), \(\alpha^2 - 2\) is also a root of the equation.</p>
Step-by-Step Solution
Key Concept: If α is a root satisfying the given condition, then α² - 2 must also be a root of the same equation. Use Vieta's formulas by considering what constraints this self-mapping property places on the coefficients a and b.
<p><strong>Step 1:</strong> Let the roots be α and β. The condition states: if α is a root, then α² - 2 is also a root.</p><p><strong>Step 2:</strong> This means either:</p><ul><li><strong>Case 1:</strong> α² - 2 = α (root maps to itself)</li><li><strong>Case 2:</strong> α² - 2 = β and β² - 2 = α (roots form a 2-cycle)</li></ul><p><strong>Step 3 (Case 1):</strong> If α² - 2 = α, then α² - α - 2 = 0, giving α = 2 or α = -1.</p><ul><li>If both roots are 2: a = -4, b = 4</li><li>If both roots are -1: a = 2, b = 1</li><li>If roots are 2 and -1: a = -1, b = -2</li></ul><p><strong>Step 4 (Case 2):</strong> If α² - 2 = β and β² - 2 = α, then α² + α - 2 = β² + β - 2, so (α² - β²) + (α - β) = 0.</p><p>This gives (α - β)(α + β + 1) = 0. Since α ≠ β, we have β = -α - 1.</p><p>Substituting into β² - 2 = α: (-α - 1)² - 2 = α gives α² + α - 1 = 0, so α = (-1 ± √5)/2.</p><ul><li>For α = (-1 + √5)/2: β = (-1 - √5)/2, giving a = 1, b = -1</li><li>For α = (-1 - √5)/2: β = (-1 + √5)/2, giving a = 1, b = -1 (same pair)</li></ul><p><strong>Step 5:</strong> Verify all pairs yield real coefficients and satisfy the original condition.</p><p>∴ Answer: <strong>4 pairs</strong> — (a,b) = (-4, 4), (2, 1), (-1, -2), (1, -1)</p>
Correct Answer: A