Let $\lambda \neq 0$ be a real number. Let $\alpha, \beta$ be the roots of the equation $14x^2 - 31x + 3\lambda = 0$ and $\alpha, \gamma$ be the roots of the equation $35x^2 - 53x + 4\lambda = 0$. Then $\dfrac{3\alpha}{\beta}$ and $\dfrac{4\alpha}{\gamma}$ are the roots of the equation:
Step-by-Step Solution
Key Concept: Use Vieta's formulas for both equations to express $\alpha+\beta$, $\alpha\beta$, $\alpha+\gamma$, $\alpha\gamma$ in terms of $\lambda$, then solve for $\alpha, \beta, \gamma, \lambda$ and compute the sum and product of $\frac{3\alpha}{\beta}$ and $\frac{4\alpha}{\gamma}$.
From Vieta's: $\alpha+\beta=\frac{31}{14}$, $\alpha\beta=\frac{3\lambda}{14}$, $\alpha+\gamma=\frac{53}{35}$, $\alpha\gamma=\frac{4\lambda}{35}$. Solving gives $\alpha=\frac{5}{7}$, $\beta=\frac{3}{2}$, $\gamma=\frac{4}{5}$, $\lambda=5$. Sum $= \frac{3\alpha}{\beta}+\frac{4\alpha}{\gamma} = \frac{10}{7}+\frac{25}{7}=5$. Product $= \frac{12\alpha^2}{\beta\gamma}=\frac{50}{49}$. Equation: $x^2-5x+\frac{50}{49}=0 \Rightarrow 49x^2-245x+250=0$.
Correct Answer: 3