Question

If $\alpha ,\beta$ are the roots of $\lambda (x^2+x)+x+5=0$ and ${\lambda }_1,{\lambda }_2$ are two values of $\lambda$ for which $\alpha ,\beta$ are connected by the relation $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=4,$ then the value of $\frac{{\lambda }_1}{{\lambda }_2}+\frac{{\lambda }_2}{{\lambda }_1}$ is equal to

Answer 1

Given equation is
$\lambda (x^2+x)+x+5=0$
$\Rightarrow \lambda x^2+(\lambda +1)x+5=0$
Roots are $\alpha ,\beta .$
Here, $\alpha +\beta =-\frac{\lambda +1}{\lambda },\alpha \beta =\frac{5}{\lambda }$

$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=4$
$\Rightarrow {\alpha }^2+{\beta }^2=4\alpha \beta$
$\Rightarrow {(\alpha +\beta )}^2=6\alpha \beta$
$\Rightarrow {\left(\frac{1+\lambda }{\lambda }\right)}^2=6\times \frac{5}{\lambda }$
$\Rightarrow \frac{1+{\lambda }^2+2\lambda }{{\lambda }^2}=\frac{30}{\lambda }$
$\Rightarrow 1+{\lambda }^2+2\lambda =30\lambda$
$\Rightarrow {\lambda }^2-28\lambda +1=0$
$\Rightarrow {\lambda }_1+{\lambda }_2=28,{\lambda }_1{\lambda }_2=1$

Now, $\frac{{\lambda }_1}{{\lambda }_2}+\frac{{\lambda }_2}{{\lambda }_1}=\frac{{\lambda }_1^2+{\lambda }_2^2}{{\lambda }_1{\lambda }_2}=\frac{{({\lambda }_1+{\lambda }_2)}^2-2{\lambda }_1{\lambda }_2}{{\lambda }_1{\lambda }_2}$
$=\frac{{\left(28\right)}^2-2\times 1}{1}=782$