Question

If $\alpha$ and $\beta$ are the zeroes of the equation $6x^2+x-2=0$, Find $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$.

Answer 1

$\alpha$ and $\beta$ are xwero of the equation $6x^2+x-2=0$

$\Rightarrow x=\frac{-1\pm \sqrt{1+48}}{12}=\frac{-1\pm 7}{12}$

$=\frac{1}{2}$ and $\frac{-2}{3}$

$\therefore \alpha =\frac{1}{2}$ and $\beta =\frac{-2}{3}$

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

$=\frac{1}{2}\times \frac{-3}{2}+\left(\frac{-2}{3}\right)\times \frac{2}{1}$

$=-\frac{3}{4}-\frac{4}{3}$

$=\frac{-9-16}{12}=\frac{-25}{12}$

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