Question

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

• A. $7 x^2+245 x-250=0$
• B. $49 x^2+245 x+250=0$
• C. $49 x^2-245 x+250=0$
• D. $7 x^2-245 x+250=0$

Answer 1

Answer: (C)

$ 14 x^2-31 x+3 \lambda=0$
$ \alpha+\beta=\frac{31}{14} \ldots . .(1) $
and $ \alpha \beta=\frac{3 \lambda}{14} \ldots .(2) $
$ 35 x^2-53 x+4 \lambda=0$
$ \alpha+\gamma=\frac{53}{35} \ldots .(3) $
and $ \alpha \gamma=\frac{4 \lambda}{35} \ldots .(4) $
$ \frac{(2)}{(4)} \Rightarrow \frac{\beta}{\gamma}=\frac{3 \times 35}{4 \times 14}=\frac{15}{8} \Rightarrow \beta=\frac{15}{8} \gamma$
$(1)-(3) \Rightarrow \beta-\gamma=\frac{31}{14}-\frac{53}{35}=\frac{155-106}{70}=\frac{7}{10}$
$ \frac{15}{8} \gamma-\gamma=\frac{7}{10} \Rightarrow \gamma=\frac{4}{5} $
$\Rightarrow \beta=\frac{15}{8} \times \frac{4}{5}=\frac{3}{2} $
$ \Rightarrow \alpha=\frac{31}{14}-\beta=\frac{31}{14}-\frac{3}{2}=\frac{5}{7} $
$ \Rightarrow \lambda=\frac{14}{3} \alpha \beta=\frac{14}{3} \times \frac{5}{7} \times \frac{3}{2}=5$
so, sum of roots $\frac{3 \alpha}{\beta}+\frac{4 \alpha}{\gamma}=\left(\frac{3 \alpha \gamma+4 \alpha \beta}{\beta \gamma}\right)$
$=\frac{\left(3 \times \frac{4 \lambda}{35}+4 \times \frac{3 \lambda}{14}\right)}{\beta \gamma}=\frac{12 \lambda(14+35)}{14 \times 35 \beta \gamma}$
$=\frac{49 \times 12 \times 5}{490 \times \frac{3}{2} \times \frac{4}{5}}=5$
Product of roots
$=\frac{3 \alpha}{\beta} \times \frac{4 \alpha}{\gamma}=\frac{12 \alpha^2}{\beta \gamma}=\frac{12 \times \frac{25}{49}}{\frac{3}{2} \times \frac{4}{5}}=\frac{250}{49}$
So, required equation is $x^2-5 x+\frac{250}{49}=0$
$\Rightarrow 49 x^2-245 x+250=0$