Question

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

• A. $-\frac{25}{12}$
• B. $\frac{25}{12}$
• C. $\frac{25}{36}$
• D. $\frac{12}{25}$

Answer 1

The correct option is A

$-\frac{25}{12}$

Given polynomial is :
$6x^2+x-2a{x}^2+bx+c\Rightarrow a=6;b=1;c=-2\alpha +\beta =\frac{-b}{a}=\frac{-1}{6}\alpha \beta =\frac{c}{a}=\frac{-2}{6}=\frac{-1}{3}\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }$

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