Answer :
To determine the values of \( x \) for which the quotient
[tex]\[ \frac{\sqrt{6(x-1)}}{\sqrt{3x^2}} \][/tex]
is defined, we need to analyze the conditions under which both the numerator and the denominator are defined, as well as when the denominator is non-zero.
1. Numerator Condition:
[tex]\[ \sqrt{6(x-1)} \][/tex]
For the square root function to be defined, the expression inside the square root must be non-negative:
[tex]\[ 6(x-1) \geq 0 \][/tex]
Dividing both sides by 6, we get:
[tex]\[ x - 1 \geq 0 \][/tex]
Adding 1 to both sides, we find:
[tex]\[ x \geq 1 \][/tex]
2. Denominator Condition:
[tex]\[ \sqrt{3x^2} \][/tex]
The square root function itself is defined for all real values \( x \) because \( x^2 \) is always non-negative. However, since this expression is in the denominator, it must be non-zero:
[tex]\[ \sqrt{3x^2} \neq 0 \][/tex]
Simplifying inside the square root, we get:
[tex]\[ 3x^2 \neq 0 \][/tex]
This implies:
[tex]\[ x \neq 0 \][/tex]
3. Combined Conditions:
We need both conditions to be satisfied simultaneously:
- The numerator \(\sqrt{6(x-1)}\) requires \( x \geq 1 \).
- The denominator \(\sqrt{3x^2}\) requires \( x \neq 0 \).
Considering both conditions together, \( x = 0 \) is excluded from the valid range, but since \( x \geq 1 \) already excludes \( x = 0 \), the final condition is:
[tex]\[ x \geq 1 \][/tex]
Hence, the inequality representing all values of \( x \) for which the quotient \(\frac{\sqrt{6(x-1)}}{\sqrt{3x^2}}\) is defined is:
[tex]\[ \boxed{x \geq 1} \][/tex]
Which corresponds to option C.
[tex]\[ \frac{\sqrt{6(x-1)}}{\sqrt{3x^2}} \][/tex]
is defined, we need to analyze the conditions under which both the numerator and the denominator are defined, as well as when the denominator is non-zero.
1. Numerator Condition:
[tex]\[ \sqrt{6(x-1)} \][/tex]
For the square root function to be defined, the expression inside the square root must be non-negative:
[tex]\[ 6(x-1) \geq 0 \][/tex]
Dividing both sides by 6, we get:
[tex]\[ x - 1 \geq 0 \][/tex]
Adding 1 to both sides, we find:
[tex]\[ x \geq 1 \][/tex]
2. Denominator Condition:
[tex]\[ \sqrt{3x^2} \][/tex]
The square root function itself is defined for all real values \( x \) because \( x^2 \) is always non-negative. However, since this expression is in the denominator, it must be non-zero:
[tex]\[ \sqrt{3x^2} \neq 0 \][/tex]
Simplifying inside the square root, we get:
[tex]\[ 3x^2 \neq 0 \][/tex]
This implies:
[tex]\[ x \neq 0 \][/tex]
3. Combined Conditions:
We need both conditions to be satisfied simultaneously:
- The numerator \(\sqrt{6(x-1)}\) requires \( x \geq 1 \).
- The denominator \(\sqrt{3x^2}\) requires \( x \neq 0 \).
Considering both conditions together, \( x = 0 \) is excluded from the valid range, but since \( x \geq 1 \) already excludes \( x = 0 \), the final condition is:
[tex]\[ x \geq 1 \][/tex]
Hence, the inequality representing all values of \( x \) for which the quotient \(\frac{\sqrt{6(x-1)}}{\sqrt{3x^2}}\) is defined is:
[tex]\[ \boxed{x \geq 1} \][/tex]
Which corresponds to option C.
