11. For the function $f(x)=\sqrt{a x^{2}+b x}$, there exists a positive number $b$, such that the domain and range of $f(x)$ are the same. Then the value of the non-zero real number $a$ is $\qquad$.
11. -4 .

If $a>0$, for the positive number $b$, the domain of $f(x)$ is
$$
D=\left(-\infty,-\frac{b}{a}\right] \cup[0,+\infty) .
$$

However, the range of $f(x)$, $A \subseteq[0,+\infty)$, so $D \neq A$, which does not meet the requirement.
If $a>0$, so, $a=-4$.