Question:
If $\alpha, \beta$ are the zeros of the polynomial $x^{2}+6 x+2$, then $\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=?$
(a) 3
(b) −3
(c) 12
(d) −12
Solution:
(b) $-3$
Since $\alpha$ and $\beta$ are the zeroes of $x^{2}+6 x+2$, we have:
$\alpha+\beta=-6$ and $\alpha \beta=2$
$\therefore\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=\left(\frac{\alpha+\beta}{\alpha \beta}\right)=\frac{-6}{2}=-3$