Question

Let $\lambda \ne 0$ 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 $\frac{3\alpha }{\beta }$ and $\frac{4\alpha }{\gamma }$ are the roots of the equation

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

Answer 1

Answer: (C)

$14x^2-31x+3\lambda =0$
$\alpha +\beta =\frac{31}{14}\dots ..(1)$
and $\alpha \beta =\frac{3\lambda }{14}\dots .(2)$
$35x^2-53x+4\lambda =0$
$\alpha +\gamma =\frac{53}{35}\dots .(3)$
and $\alpha \gamma =\frac{4\lambda }{35}\dots .(4)$
$\frac{(2)}{(4)}\Rightarrow \frac{\beta }{\gamma }=\frac{3\times 35}{4\times 14}=\frac{15}{8}\Rightarrow \beta =\frac{15}{8}\gamma$
$(1)-(3)\Rightarrow \beta -\gamma =\frac{31}{14}-\frac{53}{35}=\frac{155-106}{70}=\frac{7}{10}$
$\frac{15}{8}\gamma -\gamma =\frac{7}{10}\Rightarrow \gamma =\frac{4}{5}$
$\Rightarrow \beta =\frac{15}{8}\times \frac{4}{5}=\frac{3}{2}$
$\Rightarrow \alpha =\frac{31}{14}-\beta =\frac{31}{14}-\frac{3}{2}=\frac{5}{7}$
$\Rightarrow \lambda =\frac{14}{3}\alpha \beta =\frac{14}{3}\times \frac{5}{7}\times \frac{3}{2}=5$
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 $x^2-5x+\frac{250}{49}=0$
$\Rightarrow 49x^2-245x+250=0$