If $\alpha ,\beta$ are the roots of the equation $\lambda \left(x^2-x\right)+x+5=0$ and if ${\lambda }_1,{\lambda }_2$ are two values of $\lambda$ for which  the roots $\alpha ,\beta$ are related by $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{4}{5}$ then if the value of $\frac{{\lambda }_1}{{\lambda }_2}+\frac{{\lambda }_2}{{\lambda }_1}=255-k^2$ then |k|=  

Given equation  is  $\lambda \left(x^2-x\right)+x+5=0$

$\Rightarrow \lambda x^2-\lambda x+x+5=0$

$\Rightarrow \lambda x^2-x\left(\lambda -1\right)+5=0$

Roots are $\alpha ,\beta$

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

Given   $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{4}{5}\Rightarrow \frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{4}{5}$

$\Rightarrow 5\left({\alpha }^2+{\beta }^2\right)=4\alpha \beta$

$\Rightarrow 5\left({\alpha }^2+{\beta }^2+2\alpha \beta \right)=14\alpha \beta$

$\Rightarrow 5{\left(\alpha +\beta \right)}^2=14\alpha \beta$

$\Rightarrow 5{\left(\frac{\lambda -1}{\lambda }\right)}^2=14\left(\frac{5}{\lambda }\right)$

$\Rightarrow \frac{{\left(\lambda -1\right)}^2}{{\lambda }^2}=\frac{14}{\lambda }$

$\Rightarrow {\lambda }^2+1-2\lambda =14\lambda$

$\Rightarrow {\lambda }^2-16\lambda +1=0$

Roots are     ${\lambda }_1,{\lambda }_2$

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

given          $\frac{{\lambda }_1}{{\lambda }_2}+\frac{{\lambda }_2}{{\lambda }_1}=255-k^2$

$\Rightarrow \frac{{{\lambda }_1}^2+{{\lambda }_2}^2}{{\lambda }_1{\lambda }_2}=255-k^2$

$\Rightarrow \frac{{\left({\lambda }_1+{\lambda }_2\right)}^2-2{\lambda }_1{\lambda }_2}{{\lambda }_1{\lambda }_2}=255-k^2$

$\Rightarrow \frac{256-2}{1}=255-K^2$

$\Rightarrow 254=255-K^2\Rightarrow K^2=1\Rightarrow K=\pm 1$

$\left|K\right|=1$