Let α, β be the roots of the quadratic equation x²+√6 x+3=0. Then (α²³+β²³+α¹⁴+β¹⁴)/(α¹⁵+β¹⁵+α¹⁰+β¹⁰) is equal to :

Solution

Maths sir · 4 steps

Find a small power of α that is a plain number. Here α⁴ = −9, and the big powers collapse.

1. Find a clean power. From the equation, α² + 3 = -√6 α. Square both sides to get α⁴ + 6α² + 9 = 6α², so α⁴ = -9. The same holds for β, so α⁸ = β⁸ = 81.
2. Rewrite the numerator. Use α²³ = α⁸ · α¹⁵ = 81α¹⁵ and α¹⁴ = α⁴ · α¹⁰ = -9α¹⁰, and do the same for β. Write Sₙ = αⁿ + βⁿ.

   (S₂₃ + S₁₄)/(S₁₅ + S₁₀) = (81 S₁₅ - 9 S₁₀)/(S₁₅ + S₁₀)

3. S10 vanishes. α¹⁰ = α⁸ · α² = 81α², so S₁₀ = 81(α² + β²) = 81[(α+β)² - 2αβ] = 81(6 - 6) = 0. Also S₁₅ = -729 S₃ and S₃ = (α+β)³ - 3αβ(α+β) = -6√6 + 9√6 = 3√6. So S₁₅ ≠ 0 and it can be cancelled.
4. So. The expression is (81 S₁₅)/S₁₅ = 81, which is option (d).

The idea to keep. When you see huge powers of roots, look for the smallest power that becomes a real number, here α⁴ = -9. Then reduce every power with it and check which power sums become zero.

Answer from the NTA final answer key. Solution by PrepWiser.