**Question:**

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

**Solution:**

By using the relationship between the zeroes of the quadratic ploynomial.

We have,

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

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

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

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

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

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

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

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