JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let \(\lambda \neq 0\) be in \(R\). If \(\alpha\) and \(\beta\) are the roots of the equation, \(x^{2}-x+2 \lambda=0\) and \(\alpha\) and \(\gamma\) are the roots of the equation, \(3 x^{2}-10 x+27 \lambda=0\) then \(\frac{\beta \gamma}{\lambda}\) is equal to
- A \(36\)
- B \(27\)
- C \(9\)
- D \(18\)
Answer & Solution
Correct Answer
(D) \(18\)
Step-by-step Solution
Detailed explanation
\(\alpha+\beta=1, \alpha \beta=2 \lambda\) \(\alpha+\beta=\frac{10}{3}, \quad \alpha \gamma=\frac{27 \lambda}{3}=9 \lambda\) \(\gamma-\beta=\frac{7}{3}\) \(\frac{\gamma}{\beta}=\frac{9}{2} \Rightarrow \gamma=\frac{9}{2} \beta=\frac{9}{2} \times \frac{2}{3} \Rightarrow \gamma=3\)…
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