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(C) For the equation $14 x^2-31 x+3 \lambda=0$,we have $\alpha+\beta=\frac{31}{14}$ and $\alpha \beta=\frac{3 \lambda}{14}$.For the equation $35 x^2-5

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$49 x^2-245 x+250=0$

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(C) For the equation $14 x^2-31 x+3 \lambda=0$,we have $\alpha+\beta=\frac{31}{14}$ and $\alpha \beta=\frac{3 \lambda}{14}$.For the equation $35 x^2-53 x+4 \lambda=0$,we have $\alpha+\gamma=\frac{53}{35}$ and $\alpha \gamma=\frac{4 \lambda}{35}$.Subtracting the two equations for $\alpha$: $(\alpha+\beta)-(\alpha+\gamma) = \frac{31}{14}-\frac{53}{35}$ $\Rightarrow \beta-\gamma = \frac{155-106}{70} = \frac{49}{70} = \frac{7}{10}$.From $\alpha \beta = \frac{3 \lambda}{14}$ and $\alpha \gamma = \frac{4 \lambda}{35}$,we get $\frac{\beta}{\gamma} = \frac{3 \lambda}{14} 	imes \frac{35}{4 \lambda} = \frac{15}{8}$,so $\beta = \frac{15}{8} \gamma$.Substituting $\beta$ in $\beta-\gamma = \frac{7}{10}$: $\frac{15}{8} \gamma - \gamma = \frac{7}{10}$ $\Rightarrow \frac{7}{8} \gamma = \frac{7}{10}$ $\Rightarrow \gamma = \frac{4}{5}$.Then $\beta = \frac{15}{8} 	imes \frac{4}{5} = \frac{3}{2}$ and $\alpha = \frac{31}{14} - \frac{3}{2} = \frac{31-21}{14} = \frac{10}{14} = \frac{5}{7}$.Now,$\lambda = \frac{14 \alpha \beta}{3} = \frac{14}{3} 	imes \frac{5}{7} 	imes \frac{3}{2} = 5$.The roots of the required equation are $x_1 = \frac{3 \alpha}{\beta} = \frac{3(5/7)}{3/2} = \frac{15/7}{3/2} = \frac{10}{7}$ and $x_2 = \frac{4 \alpha}{\gamma} = \frac{4(5/7)}{4/5} = \frac{20/7}{4/5} = \frac{25}{7}$.Sum of roots: $x_1+x_2 = \frac{10}{7} + \frac{25}{7} = \frac{35}{7} = 5$.Product of roots: $x_1 x_2 = \frac{10}{7} 	imes \frac{25}{7} = \frac{250}{49}$.The required equation is $x^2 - (x_1+x_2)x + x_1 x_2 = 0$ $\Rightarrow x^2 - 5x + \frac{250}{49} = 0$ $\Rightarrow 49 x^2 - 245 x + 250 = 0$.

Let $\lambda \neq 0$ be a real number. Let $\alpha, \beta$ be the roots of the equation $14 x^2-31 x+3 \lambda=0$ and $\alpha, \gamma$ be the roots of the equation $35 x^2-53 x+4 \lambda=0$. Then $\frac{3 \alpha}{\beta}$ and $\frac{4 \alpha}{\gamma}$ are the roots of the equation :

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