Let [ 0] be a real number.

Question

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

Accepted Answer

$14 x^{2}-31 x+3 \lambda=0$

$$
\begin{aligned}
& \alpha+\beta=\frac{31}{14} \ldots .(1) \text { and } \alpha \beta=\frac{3 \lambda}{14}\quad...(2) \\
& 35 x^{2}-53 x+4 \lambda=0 \\
& \alpha+\gamma=\frac{53}{35} \ldots(3) \text { and } \alpha \gamma=\frac{4 \lambda}{35} \quad\ldots(4) \\
& \frac{(2)}{(4)} \Rightarrow \frac{\beta}{\gamma}=\frac{3 \times 35}{4 \times 14}=\frac{15}{8} \Rightarrow \beta=\frac{15}{8} \gamma
\end{aligned}
$$

(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 $\frac{3 \alpha}{\beta}+\frac{4 \alpha}{\gamma}=\left(\frac{3 \alpha \gamma+4 \alpha \beta}{\beta \gamma}\right)$

$$
\begin{aligned}
& =\frac{\left(3 \times \frac{4 \lambda}{35}+4 \times \frac{3 \lambda}{14}\right)}{\beta \gamma}=\frac{12 \lambda(14+35)}{14 \times 35 \beta \gamma} \\
& =\frac{49 \times 12 \times 5}{490 \times \frac{3}{2} \times \frac{4}{5}}=5
\end{aligned}
$$

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 $x^{2}-5 x+\frac{250}{49}=0$

$\Rightarrow 49 \mathrm{x}^{2}-245 \mathrm{x}+250=0$

Solutions · 1