If 3 and $1 / 3$ are the roots of the equation $a x^{2}-6 x+c=0$, what is the value of $a+c$?
(a) 1
(b) 0
(c) $-\frac{9}{5}$
(d) $\frac{18}{5}$
(e) -5
The correct option is (d).

Solution 1: Since 3 and $\frac{1}{3}$ are roots of the equation $a x^{2}-6 x+c=0$, we have $9 a-18+c=0$ and $\frac{1}{9} a-2+c=0$, that is, $9 a+c=18$ and $a+9 c=18$. Adding these two equations, we get $10(a+c)=10 a+10 c=36$, that is, $a+c=36 / 10=18 / 5$.

Solution 2: In a quadratic equation $a x^{2}+b x+c=0$, the sum of the roots is $-b / a$ and the product is $c / a$. Since $b=-6$, we get $10 / 3=3+\frac{1}{3}=6 / a$ and $1=3 \times \frac{1}{3}=c / a$, that is, $a=c=9 / 5$. Thus, $a+c=18 / 5$.