<p>If α, β, γ are the roots of \(x^3 + 2x^2 - 3x - 1 = 0\), then \(α^{-2} + β^{-2} + γ^{-2}\) is equal to</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to find the elementary symmetric functions of the roots, then express α⁻² + β⁻² + γ⁻² in terms of these symmetric functions using algebraic identities.
<p><strong>Step 1: Apply Vieta's Formulas</strong></p><p>For the polynomial x³ + 2x² - 3x - 1 = 0 with roots α, β, γ:</p><p>• α + β + γ = -2</p><p>• αβ + βγ + γα = -3</p><p>• αβγ = 1</p><p><strong>Step 2: Express α⁻² + β⁻² + γ⁻²</strong></p><p>α⁻² + β⁻² + γ⁻² = 1/α² + 1/β² + 1/γ²</p><p>Finding a common denominator:</p><p>= (β²γ² + α²γ² + α²β²)/(α²β²γ²)</p><p><strong>Step 3: Find the denominator</strong></p><p>Since αβγ = 1, we have (αβγ)² = 1</p><p>Therefore: α²β²γ² = 1</p><p><strong>Step 4: Find the numerator β²γ² + α²γ² + α²β²</strong></p><p>We use the identity:</p><p>(αβ + βγ + γα)² = α²β² + β²γ² + γ²α² + 2αβγ(α + β + γ)</p><p>Substituting known values:</p><p>(-3)² = α²β² + β²γ² + γ²α² + 2(1)(-2)</p><p>9 = α²β² + β²γ² + γ²α² - 4</p><p>α²β² + β²γ² + γ²α² = 13</p><p><strong>Step 5: Calculate the final answer</strong></p><p>α⁻² + β⁻² + γ⁻² = 13/1 = 13</p><p><strong>∴ Answer: b</strong></p>
Correct Answer: b