Question

$\alpha ,\beta$ are the root of the equation $\lambda (x^2-x)+x+5=0.$ If ${\lambda }_{1}and{\lambda }_2$ are the two values of $\lambda$ for which the roots $\alpha ,\beta$ are connected by the relation $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{4}{5},$ find the value of $\frac{{\lambda }_1}{{\lambda }_2}+\frac{{\lambda }_2}{{\lambda }_1}.$

Answer 1

The given equation is $\lambda x^2-(\lambda -1)x+5=0.$
$\therefore \alpha +\beta =\frac{\lambda -1}{\lambda },\alpha \beta =\frac{5}{\lambda }$
Also
$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{4}{5}or5({\alpha }^2+{\beta }^2)=4\alpha \beta$
or $5[{(\alpha +\beta )}^2-2\alpha \beta ]=4\alpha \beta$
or $5{\left(\frac{\lambda -1}{\lambda }\right)}^2=14.\frac{5}{\lambda }$
or ${\lambda }^2-2\lambda +1=14\lambda or{\lambda }^2-16\lambda +1=0$
Its roots are ${\lambda }_1,{\lambda }_2$
$\therefore {\lambda }_1+{\lambda }_2=16,{\lambda }_1{\lambda }_2=1$
Now
$\frac{{\lambda }_1}{{\lambda }_2}+\frac{{\lambda }_2}{{\lambda }_1}=\frac{{{\lambda }_1}^2+{{\lambda }_2}^2}{{\lambda }_1{\lambda }_2}$
$=\frac{{({\lambda }_1+{\lambda }_2)}^2-2{\lambda }_1{\lambda }_2}{{\lambda }_1{\lambda }_2}=256-2=254.$