$\alpha ,\beta$ are the roots of the equation  $\lambda$(x² - x) + x+5 = 0 .If  $\lambda$₁ and $\lambda$₂ be the two values of $\lambda$ , determined from the equation $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{4}{5}$, then  the value of $\frac{{\lambda }_1}{{\lambda }_2}+\frac{{\lambda }_2}{{\lambda }_1}=$

The given equation reduces to

$\lambda x^2+(1-\lambda )x+5=0$ 

Let its roots are $\alpha$ and $\beta$

$\alpha +\beta =\frac{\lambda -1}{\lambda },\alpha \beta =\frac{5}{\lambda }$

Now,

$\begin{array}{l}\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{4}{5}\\ \frac{{\alpha }^2+b^2}{\alpha \beta }=\frac{4}{5}\\ \frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }=\frac{4}{5}\\ \frac{(\lambda -1{)}^2}{{\lambda }^2}-\frac{10}{\lambda }=\frac{4}{5}\\ \frac{(\lambda -1{)}^2-10\lambda }{5\lambda }=\frac{4}{5}\\ {\lambda }^2-16\lambda +1=0\end{array}$

Let its roots are  $\lambda$₁ and $\lambda$₂

Thus, $,{\lambda }_1+{\lambda }_2=16,{\lambda }_1{\lambda }_2\stackrel{ }{=}1$

$\begin{array}{l}\frac{{\lambda }_1}{{\lambda }_2}+\frac{{\lambda }_2}{{\lambda }_1}\\ =\frac{{\left({\lambda }_1+{\lambda }_2\right)}^2-2{\lambda }_1{\lambda }_2}{{\lambda }_1{\lambda }_2}\\ =\frac{256-2}{1}\\ =254\end{array}$