jmc

Since $x=a$ is a root of $x^2+ax+b=0,$ $$a^2+a^2+b=0,$$or $2a^2+b=0,$ so $b=-2a^2.$

Since $x=b$ is a root of $x^2+ax+b=0,$ $$b^2+ab+b=0.$$This factors as $b(b+a+1)=0,$ so $b=0$ or $a+b+1=0.$

If $b=0,$ then $-2a^2=0,$ so $a=0.$

If $a+b+1=0,$ then $-2a^2+a+1=0.$ This equation factors as $-(a-1)(2a+1)=0,$ so $a=1$ or $a=-\frac{1}{2}.$ If $a=1,$ then $b=-2.$ If $a=-\frac{1}{2},$ then $b=-\frac{1}{2}.$

Therefore, there are $\boxed{3}$ ordered pairs $(a,b),$ namely $(0,0),$ $(1,-2),$ and $\left(-\frac{1}{2},-\frac{1}{2}\right).$