JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
If the sum of the squares of the reciprocals of the roots \(\alpha\) and \(\beta\) of the equation \(3 x^{2}+\lambda x-1=0\) is 15 , then \(6\left(\alpha^{3}+\beta^{3}\right)^{2}\) is equal to
- A \(18\)
- B \(24\)
- C \(36\)
- D \(96\)
Answer & Solution
Correct Answer
(B) \(24\)
Step-by-step Solution
Detailed explanation
Here \(\alpha, \beta\) roots of equation \(3 x^{2}+\lambda x-1=0\) \(\alpha+\beta=\frac{-\lambda}{3}, \alpha \beta=\frac{-1}{3}\) \(\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha^{2} \beta^{2}}=15\) \(\lambda^{2}=9\) Now…
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