If α and β are the zeros of the quadratic polynomial

Solution:

Since $\alpha$ and $\beta$ are the zeros of the quadratics polynomial

$f(x)=6 x^{2}+x-2$

sum of zeros $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$\alpha+\beta=-\frac{1}{6}$

Product of the zeros $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

$\alpha \beta=-\frac{1}{3}$

We have, $\frac{\alpha+\beta}{\beta+\alpha}$

$\frac{\alpha+\beta}{\beta+\alpha}=\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}$

$\frac{\alpha+\beta}{\beta+\alpha}=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha \beta}$

By substituting $\alpha+\beta=\frac{-1}{6}$ and $\alpha \beta=-\frac{1}{3}$ we get,

$\frac{\alpha+\beta}{\beta+\alpha}=\frac{\left(-\frac{1}{6}\right)^{2}-2\left(-\frac{1}{3}\right)}{-\frac{1}{3}}$

$\frac{\alpha+\beta}{\beta+\alpha}=\frac{\frac{1}{36}+\frac{2}{3}}{\frac{-1}{3}}$

$\frac{\alpha+\beta}{\beta+\alpha}=\frac{\frac{1}{36}+\frac{24}{36}}{\frac{-1}{3}}$

$\frac{\alpha+\beta}{\beta+\alpha}=\frac{\frac{25}{36}}{-\frac{1}{3}}$

$\frac{\alpha+\beta}{\beta+\alpha}=\frac{-25}{12}$

Hence, the value of $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$ is $\frac{-25}{12}$.