Mathematics - Quadratic Equation Question with Solution | TestHub

Given $\alpha$ and $\beta$ are roots of the below equation

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

So, $\alpha +\beta =\frac{31}{14} \dots \left(1\right)$

And, $\alpha \beta =\frac{3\lambda }{14} \dots \left(2\right)$

Given $\alpha$ and $\gamma$ are roots of the below equation

$35x^2-53x+4\lambda =0$

So, $\alpha +\gamma =\frac{53}{35} \dots \left(3\right)$

And, $\alpha \gamma =\frac{4\lambda }{35} \dots \left(4\right)$

On dividing equation$\left(2\right)$  and equation $\left(4\right)$,

$\frac{\beta }{\gamma }=\frac{3\times 35}{4\times 14}=\frac{15}{8}\Rightarrow \beta =\frac{15}{8}\gamma$

On subtracting equation $\left(1\right)$ and equation $\left(3\right)$,

$\Rightarrow \beta -\gamma =\frac{31}{14}-\frac{53}{35}=\frac{155-106}{70}=\frac{7}{10}$

$\Rightarrow \frac{15}{8}\gamma -\gamma =\frac{7}{10}\Rightarrow \gamma =\frac{4}{5}$

$\Rightarrow \beta =\frac{15}{8}\times \frac{4}{5}=\frac{3}{2}$

So, $\alpha =\frac{31}{14}-\beta =\frac{31}{14}-\frac{3}{2}=\frac{5}{7}$

And $\lambda =\frac{14}{3}\alpha \beta =\frac{14}{3}\times \frac{5}{7}\times \frac{3}{2}=5$

So, the sum of the roots of the required equation,

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

$=\frac{\left(3\times \frac{4\lambda }{35}+4\times \frac{3\lambda }{14}\right)}{\beta \gamma }=\frac{12\lambda \left(14+35\right)}{14\times 35\beta \gamma }$

$=\frac{49\times 12\times 5}{490\times \frac{3}{2}\times \frac{4}{5}}=5$

And the product of the roots of the required equation,
$=\frac{3\alpha }{\beta }\times \frac{4\alpha }{\gamma }=\frac{12{\alpha }^2}{\beta \gamma }=\frac{12\times \frac{25}{49}}{\frac{3}{2}\times \frac{4}{5}}=\frac{250}{49}$

So, the required equation is

$x^2-\left(\text{sum of roots}\right)x+\text{product of roots}=0$

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

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