The Solution:
Given:
[tex]\sqrt{\left(x-6\right)}\times \sqrt{\left(x+3\right)}[/tex]Required:
To determine the range of values for x.
Applying the rule of radicals that states that:
[tex]\begin{gathered} \sqrt{a}\times\sqrt{b}=\sqrt{a\times b} \\ Where\text{ }a\ge0,b\ge0 \end{gathered}[/tex]We have:
[tex]\begin{gathered} \sqrt{\left(x-6\right)}\times\sqrt{\left(x+3\right)}=\sqrt{(x-6)(x+3)} \\ \\ Where \\ x-6\ge0\text{ } \\ x\ge6\text{ } \end{gathered}[/tex]Therefore, the correct answer is [option A]