jmc

We use the fact that the sum and product of the roots of a quadratic equation $x^2+ax+b=0$ are given by $-a$ and $b$, respectively.

In this problem, we see that $2a+b=-a$ and $(2a)(b)=b$. From the second equation, we see that either $2a=1$ or else $b=0$. But if $b=0$, then the first equation gives $2a=-a$, implying that $a=0$. This makes the two solutions of our original polynomial the same, and we are given that they are distinct. Hence $b\ne 0$, so $2a=1,$ or $a=1/2$. Then $b=-3a=-3/2$, so $a+b=\boxed{-1}$.