Chapter 2 - Functions - 2.2 Domain and Range - Exercises and Problems for Section 2.2 - Exercises and Problems - Page 82: 25

$a\geq 0$

We know that the domain of the function, $n(q)=\sqrt{q^2+a}$ is all real numbers. This means that $q^2+a\geq 0$. Since $q^2$ is always positive, $a$ must be non-negative. Hence $a\geq 0$.