Answer :
To determine which values of \( x \) make the expression \(\frac{5-x}{x(x-2)}\) undefined, we need to analyze the denominator \(x(x-2)\).
For any rational expression like \(\frac{5-x}{x(x-2)}\), the expression becomes undefined when the denominator is equal to zero, because division by zero is undefined.
Let's solve for \( x \) when the denominator \(x(x-2)\) is equal to zero:
[tex]\[ x(x-2) = 0 \][/tex]
This equation can be factored to:
[tex]\[ x = 0 \quad \text{or} \quad x - 2 = 0 \][/tex]
So, \(x = 0\) and \(x = 2\) are the solutions of this equation. Therefore, the expression \(\frac{5-x}{x(x-2)}\) is undefined when \(x = 0\) or \(x = 2\).
Additionally, the numerator \(5 - x\) does not impact the value of \( x \) that makes the denominator zero but we need to examine all potential cases. For completeness, let's check if there's a specific \( x \) value from the numerator that might make this overall problem contextually significant:
[tex]\[ 5 - x = 0 \][/tex]
Solving for \(x\):
[tex]\[ x = 5 \][/tex]
Although the expression itself wouldn't be undefined due to the numerator being zero, in the context of the problem, \( x = 5 \) makes the box's "height" as zero (ignoring division by zero issues), indicating a potentially special case for height consideration in terms of dimensions.
Therefore, combining these aspects:
- The denominator \( x(x-2) = 0 \) gives \( x = 0 \) or \( x = 2 \)
- Contextually, \( x = 5 \) is also noteworthy for making numerator zero.
As such, \(x = 0\), \(x=2\), and \(x=5\) are the critical values.
Hence, the complete values of \(x\) that make the expression undefined are:
[tex]\[ x = 0, 2, \text{ and } 5 \][/tex]
Given the options, \(0\) and \(2\) \( (i.e., as option \( C: 0 \text{ and } 2\) ) correctly aligns with causing coefficient singularity on the denominator.
Thus, the answer to the question is Option C: 0 and 2.
For any rational expression like \(\frac{5-x}{x(x-2)}\), the expression becomes undefined when the denominator is equal to zero, because division by zero is undefined.
Let's solve for \( x \) when the denominator \(x(x-2)\) is equal to zero:
[tex]\[ x(x-2) = 0 \][/tex]
This equation can be factored to:
[tex]\[ x = 0 \quad \text{or} \quad x - 2 = 0 \][/tex]
So, \(x = 0\) and \(x = 2\) are the solutions of this equation. Therefore, the expression \(\frac{5-x}{x(x-2)}\) is undefined when \(x = 0\) or \(x = 2\).
Additionally, the numerator \(5 - x\) does not impact the value of \( x \) that makes the denominator zero but we need to examine all potential cases. For completeness, let's check if there's a specific \( x \) value from the numerator that might make this overall problem contextually significant:
[tex]\[ 5 - x = 0 \][/tex]
Solving for \(x\):
[tex]\[ x = 5 \][/tex]
Although the expression itself wouldn't be undefined due to the numerator being zero, in the context of the problem, \( x = 5 \) makes the box's "height" as zero (ignoring division by zero issues), indicating a potentially special case for height consideration in terms of dimensions.
Therefore, combining these aspects:
- The denominator \( x(x-2) = 0 \) gives \( x = 0 \) or \( x = 2 \)
- Contextually, \( x = 5 \) is also noteworthy for making numerator zero.
As such, \(x = 0\), \(x=2\), and \(x=5\) are the critical values.
Hence, the complete values of \(x\) that make the expression undefined are:
[tex]\[ x = 0, 2, \text{ and } 5 \][/tex]
Given the options, \(0\) and \(2\) \( (i.e., as option \( C: 0 \text{ and } 2\) ) correctly aligns with causing coefficient singularity on the denominator.
Thus, the answer to the question is Option C: 0 and 2.
