jmc

Consider the quadratic formula $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. In order for the quadratic to have real roots, the expression underneath the square root (the discriminant) must be either positive or equal to zero. Thus, this gives us the inequality $$\begin{array}{rl}{b}^2-4ac& \ge 0\\ \Rightarrow b^2-4(1)(16)& \ge 0\\ \Rightarrow b^2-64& \ge 0\\ \Rightarrow (b+8)(b-8)& \ge 0\end{array}$$ Thus, we find that $b\in \boxed{(-\infty ,-8]\cup [8,\infty )}$.