Question

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

Answer 1

Since $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f\left(x\right)=6x^2+x-2$,

Sum of the zeroes = $(\alpha +\beta )=\frac{-b}{a}=\frac{-1}{6}$

The product of the zeroes =-2/6=−1/3 $\alpha \beta =\frac{c}{a}=\frac{-2}{6}=\frac{-1}{3}$

Now,

$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{({\alpha }^2+{\beta }^2)}{\alpha \beta }$
(by taking LCM)

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

$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{\left[\right(\alpha +\beta {)}^2–2\alpha \beta ]}{\alpha \beta }$

By substitution the values of the sum of zeroes and products of the zeroes, we will get

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

$=\frac{[\frac{1}{36}+\frac{2}{3}]}{\left(\frac{-1}{3}\right)}$

$=\frac{\left[\frac{(1+24)}{36}\right]}{\left(\frac{-1}{3}\right)}$

$=\frac{\left(\frac{25}{36}\right)}{\left(\frac{-1}{3}\right)}$

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

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