Equation Anyone?

$\begin{aligned} \lambda (x^2-x) + x + 5 & = 0 \\ \lambda x^2- (\lambda - 1)x + 5 & = 0 \\ x^2 - \frac {\lambda -1}\lambda x + \frac 5\lambda & = 0 \end{aligned}$

Since $\alpha$ and $\beta$ are the roots of the equation, by Vieta's formula , we have: $\alpha + \beta = \dfrac {\lambda-1}\lambda$ and $\alpha \beta = \dfrac 5 \lambda$ . Now,

$\begin{aligned} \alpha + \beta & = \frac {\lambda-1}\lambda & \small \color{#3D99F6} \text{Squaring both sides} \\ (\alpha + \beta)^2 & = \left(\frac {\lambda-1}\lambda\right)^2 \\ \alpha^2 + 2\alpha \beta + \beta^2 & = \frac {\lambda^2-2\lambda +1}{\lambda^2} & \small \color{#3D99F6} \text{Dividing both sides by } \alpha \beta = \frac 5\lambda \\ \frac {\alpha^2 + 2\alpha \beta + \beta^2}{\color{#3D99F6}\alpha \beta} & = \frac {\lambda^2-2\lambda +1}{\lambda^2\cdot \color{#3D99F6} \frac 5 \lambda} \\ {\color{#3D99F6}\frac \alpha \beta} + 2 + \color{#3D99F6}\frac \beta \alpha & = \frac {\lambda^2-2\lambda +1}{5 \lambda} & \small \color{#3D99F6} \text{Given that }\frac \alpha \beta + \frac \beta \alpha = \frac 45 \\ 2 + \color{#3D99F6} \frac 45 & = \frac {\lambda^2-2\lambda +1}{5 \lambda} & \small \color{#3D99F6} \text{Rearranging} \\ \implies \lambda^2-16\lambda +1 & = 0 \end{aligned}$

By Vieta's formula, we have $\lambda_1 + \lambda_2 = 16$ and $\lambda_1\lambda_2 = 1$ and $\dfrac {\lambda_1}{\lambda_2} + \dfrac {\lambda_2}{\lambda_1} = \dfrac {(\lambda_1+ \lambda_2)^2}{\lambda_1 \lambda_2} - 2 = \dfrac {16^2}1 - 2 = \boxed{254}$ .