The domain of $ f $ is $ \{x\ |\ ax^2 + bx\ge 0\} $. If $ a=0 $, then for every positive value of $ b $, the domain and range of $ f $ are each equal to the interval $ [0,\infty) $, so $ 0 $ is a possible value of $ a $. If $ a\ne0 $, the graph of $ y=ax^2+bx $ is a parabola with $ x $-intercepts at $ x=0 $ and $ x=-b/a $. If $ a>0 $, the domain of $ f $ is $ (-\infty,-b/a] \cup [0,\infty) $, but the range of $ f $ cannot contain negative numbers. If $ a<0 $, the domain of $ f $ is $ [0,-b/a] $. The maximum value of $ f $ occurs halfway between the $ x $-intercepts, at $ x=-b/2a $, and $$ f\left(-\frac{b}{2a}\right)=\sqrt{a\left(\frac{b^2}{4a^2}\right)+b\left(-\frac{b}{2a}\right)}= \frac{b}{2\sqrt{-a}}.$$ Hence, the range of $ f $ is $ [0,b/2\sqrt{-a}] $. For the domain and range to be equal, we must have \[ -\frac{b}{a} = \frac{b}{2\sqrt{-a}}\quad \text{so} \quad 2\sqrt{-a}=-a.\]The only solution is $ a=-4 $. Thus, there are $ \boxed{2} $ possible values of $ a $, and they are $ a=0 $ and $ a=-4 $.