7) The sum of the reciprocals of the roots of $a x^{2}+b x+c=0$ (where $a, b, c \neq 0$) ̀̀
(A) $\frac{1}{a}+\frac{1}{b}$
(B) $\frac{b}{c}$
(C) $-\frac{c}{b}$
(D) $-\frac{a}{b}$
(E) $-\frac{b}{c}$.
7) The answer is (E).

If the roots are $r$ and $s$, then $r+s=-\frac{b}{a}$ and $r \cdot s=\frac{c}{a}$, thus

$$
\frac{1}{r}+\frac{1}{s}=\frac{r+s}{r \cdot s}=\frac{-b / a}{c / a}=-\frac{b}{c}
$$