If $\alpha$ and $\beta$ are the zeros of the polynomial $f(x)=6 x^{2}+x-2$ find the value of $\left(\frac{\alpha}{\beta}+\frac{\alpha}{\beta}\right)$

using the relationship between the zeroes of the quadratic polynomial.

Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text { constant term }}{\text { coefficient of } x^{2}}$

$\therefore \alpha+\beta=\frac{-1}{6}$ and $\alpha \beta=-\frac{1}{3}$

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

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

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

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

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

$=-\frac{25}{12}$