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JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations

ધારોકે \(\lambda \neq 0\) એ વાસ્તવિક સંખ્યા છે. ધારોકે \(\alpha, \beta\) એ સમીકરણ \(14 x^2-31 x+3 \lambda=0\) ના બીજ છે અને \(\alpha, \gamma\) એ સમીકરણ \(35 x^2-53 x+4 \lambda=0\) નાં બીજ છે. તો \(\frac{3 \alpha}{\beta}\) અને \(\frac{4 \alpha}{\gamma}\) એ \(..........\) સમીકરણ નાં બીજ છે.

  1. A \(7 x ^2+245 x -250=0\)
  2. B \(7 x ^2-245 x +250=0\)
  3. C \(49 x ^2-245 x +250=0\)
  4. D \(49 x^2+245 x+250=0\)
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Answer & Solution

Correct Answer

(C) \(49 x ^2-245 x +250=0\)

Step-by-step Solution

Detailed explanation

\(14 x^2-31 x+3 \lambda=0\) \(\alpha+\beta=\frac{31}{14} \ldots .(1) \text { and } \alpha \beta=\frac{3 \lambda}{14} \ldots .(2)\) \(35 x^2-53 x+4 \lambda=0\) \(\alpha+\gamma=\frac{53}{35} \ldots(3) \text { and } \alpha \gamma=\frac{4 \lambda}{35} \ldots(4)\)…
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