College Algebra, Chapter 3, 3.1, Section 3.1, Problem 56

Determine the domain of the function $G(x) = \sqrt{x^2 - 9}$
The function is not defined when the radicand is a negative value. So,

$
\begin{equation}
\begin{aligned}
x^2 - 9 &\geq 0 && \text{Difference of squares}\\
\\
(x+3)(x-3) &\geq 0
\end{aligned}
\end{equation}
$


The factors on the left hand side of the inequality are $x+3$ and $x-3$. These factors are zero when $x$ is $-3$ and $3$, respectively. These numbers divide the number line into intervals
$(-\infty, -3],[-3,3], [3,\infty)$
By using some points on the interval...



Thus, the domain of $G(x)$ is $(-\infty, -3]\bigcup[3,\infty)$