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

The given equation can be written as ${\mathrm{\lambda x}}^2-(\mathrm{\lambda }-1)\mathrm{x}+5=0$

$\mathrm{\alpha },\mathrm{\beta }$ are the roots of this equation.

$\therefore \alpha +\beta =\frac{\lambda -1}{\lambda }\text{and}\alpha \beta =\frac{5}{\lambda }$

but given 

$\begin{array}{r}\frac{\mathrm{\alpha }}{\mathrm{\beta }}+\frac{\mathrm{\beta }}{\mathrm{\alpha }}=\frac{4}{5}\\ \Rightarrow \frac{{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2}{\mathrm{\alpha \beta }}=\frac{4}{5}\end{array}$

$\begin{array}{ll}\Rightarrow & \frac{(\mathrm{\alpha }+\mathrm{\beta }{)}^2-2\mathrm{\alpha \beta }}{\mathrm{\alpha \beta }}=\frac{4}{5}\Rightarrow \frac{\frac{(\mathrm{\lambda }-1{)}^2}{{\mathrm{\lambda }}^2}-\frac{10}{\mathrm{\lambda }}}{\frac{5}{\mathrm{\lambda }}}=\frac{4}{5}\\ \Rightarrow & \frac{(\mathrm{\lambda }-1{)}^2-10\mathrm{\lambda }}{5\mathrm{\lambda }}=\frac{4}{5}\Rightarrow {\mathrm{\lambda }}^2-12\mathrm{\lambda }+1=4\mathrm{\lambda }\\ \Rightarrow & {\mathrm{\lambda }}^2-16\mathrm{\lambda }+1=0\end{array}$

It is a quadratic in $\lambda$, let roots be ${\mathrm{\lambda }}_1 \text{and }{\mathrm{\lambda }}_2$, then

$\begin{array}{rl}{\mathrm{\lambda }}_1+{\mathrm{\lambda }}_2& =16\text{and}{\mathrm{\lambda }}_1{\mathrm{\lambda }}_2=1\\ \therefore \frac{{\mathrm{\lambda }}_1}{{\mathrm{\lambda }}_2}+\frac{{\mathrm{\lambda }}_2}{{\mathrm{\lambda }}_1}& =\frac{{\mathrm{\lambda }}_1^2+{\mathrm{\lambda }}_2^2}{{\mathrm{\lambda }}_1{\mathrm{\lambda }}_2}=\frac{{({\mathrm{\lambda }}_1+{\mathrm{\lambda }}_2)}^2-2{\mathrm{\lambda }}_1{\mathrm{\lambda }}_2}{{\mathrm{\lambda }}_1{\mathrm{\lambda }}_2}\\ & =\frac{(16{)}^2-2(1)}{1}=254\end{array}$