To determine the values of [tex]\( x \)[/tex] for which the quotient [tex]\(\sqrt{6(x-1)} \div \sqrt{3x^2}\)[/tex] is defined, we need to consider the constraints given by the expressions under the square roots. Here's a step-by-step solution:
1. Identify the conditions under each square root:
- The expression [tex]\(\sqrt{6(x-1)}\)[/tex] is defined for [tex]\(6(x-1) \geq 0\)[/tex].
- The expression [tex]\(\sqrt{3x^2}\)[/tex] is defined for [tex]\(3x^2 \geq 0\)[/tex].
2. Solve each inequality:
- For [tex]\(6(x-1) \geq 0\)[/tex]:
[tex]\[
6(x-1) \geq 0
\][/tex]
Divide both sides by 6:
[tex]\[
x - 1 \geq 0
\][/tex]
Thus:
[tex]\[
x \geq 1
\][/tex]
- For [tex]\(3x^2 \geq 0\)[/tex]:
[tex]\[
3x^2 \geq 0
\][/tex]
Since [tex]\(3x^2\)[/tex] is always non-negative (because [tex]\(x^2\)[/tex] is always non-negative for all real numbers [tex]\(x\)[/tex] and multiplying it by 3 will not change its non-negativity), this inequality is always true for all [tex]\(x\)[/tex].
3. Combine the conditions:
The expression [tex]\(\sqrt{6(x-1)}\)[/tex] requires [tex]\(x \geq 1\)[/tex]. The expression [tex]\(\sqrt{3x^2}\)[/tex] places no additional restrictions on [tex]\(x\)[/tex] because it is defined for all real numbers [tex]\(x\)[/tex]. However, [tex]\(\sqrt{3x^2}\)[/tex] equals zero when [tex]\(x = 0\)[/tex], which makes the quotient undefined.
4. Observe additional constraints:
Since [tex]\(\sqrt{3x^2}\)[/tex] becomes zero when [tex]\(x = 0\)[/tex], the quotient will be undefined at [tex]\(x = 0\)[/tex]. Fortunately, [tex]\(x = 0\)[/tex] does not satisfy [tex]\(x \geq 1\)[/tex].
Therefore, the combined condition for [tex]\(x\)[/tex] under which the given expression is defined and the quotient does not become undefined is:
[tex]\[
x \geq 1
\][/tex]
Thus, the correct inequality is:
[tex]\[
\boxed{x \geq 1}
\][/tex]
So, the option that correctly represents all values of [tex]\(x\)[/tex] for which the quotient is defined is:
D. [tex]\(x \geq 1\)[/tex]
