The roots α, β of the equation 2x² + λx − 1 = 0 satisfy 1/α² + 1/β² = 20. The value of (α³ + β³)² is:
Answer and solution
Correct answer: B
Solution
Answer: B — 121
1. For the quadratic equation, the sum of the roots is:
2. α + β = −λ/2
3. The product of the roots is:
4. αβ = −1/2
5. We are given:
6. 1/α² + 1/β² = 20
7. Combining the fractions:
8. [(α + β)² − 2αβ]/(αβ)² = 20
9. Since αβ = −1/2:
10. [(α + β)² + 1] ÷ (1/4) = 20
11. So,
12. (α + β)² + 1 = 5
(α + β)² = 4
13. Now use:
14. α³ + β³ = (α + β)³ − 3αβ(α + β)
15. = (α + β)[(α + β)² − 3αβ]
16. = (α + β)[4 + 3/2]
17. Squaring:
18. (α³ + β³)² = 4 × (11/2)²
= 4 × 121/4
= 121
19. Answer: 121.
Why this method works:
Use symmetric expressions instead of solving for the individual roots. Since αβ=−1/2, neither root is zero, so the reciprocals in the question are defined. Writing s=α+β retains exactly the information needed.
Independent check and common trap:
The reciprocal-square condition fixes s²=4, allowing s=2 or −2. The requested expression is squared, so both possibilities give the same 121. Choosing a sign for λ or solving the roots explicitly adds work without affecting the answer.
Apply: Use s=α+β, p=−1/2. (s²−2p)/p²=20 gives s²=4. Required s²(s²−3p)²=4(11/2)²=121.
Check: The reciprocal-square condition fixes s²=4, allowing s=2 or −2. The requested expression is squared, so both possibilities give the same 121. Choosing a sign for λ or solving the roots explicitly adds work without affecting the answer.
1. For the quadratic equation, the sum of the roots is:
2. α + β = −λ/2
3. The product of the roots is:
4. αβ = −1/2
5. We are given:
6. 1/α² + 1/β² = 20
7. Combining the fractions:
8. [(α + β)² − 2αβ]/(αβ)² = 20
9. Since αβ = −1/2:
10. [(α + β)² + 1] ÷ (1/4) = 20
11. So,
12. (α + β)² + 1 = 5
(α + β)² = 4
13. Now use:
14. α³ + β³ = (α + β)³ − 3αβ(α + β)
15. = (α + β)[(α + β)² − 3αβ]
16. = (α + β)[4 + 3/2]
17. Squaring:
18. (α³ + β³)² = 4 × (11/2)²
= 4 × 121/4
= 121
19. Answer: 121.
Why this method works:
Use symmetric expressions instead of solving for the individual roots. Since αβ=−1/2, neither root is zero, so the reciprocals in the question are defined. Writing s=α+β retains exactly the information needed.
Independent check and common trap:
The reciprocal-square condition fixes s²=4, allowing s=2 or −2. The requested expression is squared, so both possibilities give the same 121. Choosing a sign for λ or solving the roots explicitly adds work without affecting the answer.
G Strategy shortcut
Spot: Algebra – Quadratic Equations.Apply: Use s=α+β, p=−1/2. (s²−2p)/p²=20 gives s²=4. Required s²(s²−3p)²=4(11/2)²=121.
Check: The reciprocal-square condition fixes s²=4, allowing s=2 or −2. The requested expression is squared, so both possibilities give the same 121. Choosing a sign for λ or solving the roots explicitly adds work without affecting the answer.
G Strategy: Spot: Algebra – Quadratic Equations. Apply: Use s=α+β, p=−1/2. (s²−2p)/p²=20 gives s²=4. Required s²(s²−3p)²=4(11/2)²=121. Check: The reciprocal-square condition fixes s²=4, allowing s=2 or −2. The requested expression is squared, so both possibilities give the same 121. Choosing a sign for λ or solving the roots explicitly adds work without affecting the answer.