Question

If $\alpha \&\beta$ are the zeros of $6x^2-x-2$ , then $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=?$

• A. $\frac{-25}{12}$
• B. $\frac{-12}{25}$
• C. $\frac{24}{25}$
• D. $\frac{25}{24}$

Answer 1

The correct option is A $\frac{-25}{12}$

Sum of zeros ($\alpha +\beta$) = $\frac{1}{6}$
Product of zeros $\alpha \beta$ = $\frac{-1}{3}$


$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$ =$\frac{{\left(\alpha +\beta \right)}^2-2\mathrm{\alpha \beta }}{\mathrm{\alpha \beta }}$
$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=$ $\frac{{\left(\frac{1}{6}\right)}^2-2\left(\frac{-1}{3}\right)}{\frac{-1}{3}}$= $\frac{\frac{1}{36}+\frac{2}{3}}{\frac{-1}{3}}=\frac{\frac{25}{36}}{\frac{-1}{3}}$
=$\frac{-25}{12}$