Question

If α and β are the zeroes of a polynomial f(x) = 6x² + x − 2, find the value of $\left(\frac{\alpha }{\beta }+\frac{\beta }{\alpha }\right)$

Answer 1

By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = $\frac{-\left(\mathrm{coefficient}\mathrm{of}\mathit{}x\right)}{\mathrm{coefficent}\mathrm{of}{x}^2}$ and Product of zeroes = $\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficent}\mathrm{of}{x}^2}$
$$\begin{gathered} \therefore \alpha +\beta =\frac{-1}{6}\mathrm{and}\alpha \beta =-\frac{1}{3} \\ \mathrm{Now},\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta } \\ =\frac{{\alpha }^2+{\beta }^2+2\alpha \beta -2\alpha \beta }{\alpha \beta } \end{gathered}$$
$$\begin{gathered} =\frac{{\left(\alpha +\beta \right)}^2-2\alpha \beta }{\alpha \beta } \\ =\frac{{\left({\displaystyle \frac{-1}{6}}\right)}^2-2\left(-{\displaystyle \frac{1}{3}}\right)}{-{\displaystyle \frac{1}{3}}} \\ =\frac{{\displaystyle \frac{1}{36}+\frac{2}{3}}}{-{\displaystyle \frac{1}{3}}} \\ =-\frac{{\displaystyle 25}}{12} \end{gathered}$$