To solve this problem, we need to determine the conditions under which the product [tex]\(\sqrt{x-6} \cdot \sqrt{x+3}\)[/tex] is defined. For a square root to be defined, the expression inside the square root must be non-negative.
Let's consider each term under the square roots separately:
1. [tex]\(\sqrt{x-6}\)[/tex]:
- The expression [tex]\(x-6\)[/tex] inside the square root must be non-negative.
- This leads to the inequality [tex]\(x - 6 \geq 0\)[/tex].
- Solving for [tex]\(x\)[/tex], we get [tex]\(x \geq 6\)[/tex].
2. [tex]\(\sqrt{x+3}\)[/tex]:
- The expression [tex]\(x+3\)[/tex] inside the square root must also be non-negative.
- This leads to the inequality [tex]\(x + 3 \geq 0\)[/tex].
- Solving for [tex]\(x\)[/tex], we get [tex]\(x \geq -3\)[/tex].
To ensure that both square roots are defined simultaneously, we need to find the values of [tex]\(x\)[/tex] that satisfy both conditions.
We have the inequalities:
- [tex]\(x \geq 6\)[/tex]
- [tex]\(x \geq -3\)[/tex]
The more restrictive condition is [tex]\(x \geq 6\)[/tex]. If [tex]\(x \geq 6\)[/tex], then it will automatically satisfy [tex]\(x \geq -3\)[/tex] as well.
Therefore, the inequality that represents all values of [tex]\(x\)[/tex] for which [tex]\(\sqrt{x-6} \cdot \sqrt{x+3}\)[/tex] is defined is:
[tex]\[
\boxed{x \geq 6}
\][/tex]
Thus, the correct answer is:
A. [tex]\(x \geq 6\)[/tex]
