- The excluded values of a rational expression are the values of \(x\) for which the denominator equals zero.
- For the excluded values to be \(2\) and \(5\), the denominator must be a quadratic expression of the form \(k(x-2)(x-5)\), which simplifies to \(k(x^2-7x+10)\).
- Testing the given options, we identify that the denominator \(x^2-7x+10\) evaluates to zero at exactly \(x=2\) and \(x=5\).
- Therefore, the correct rational expression is \(\boxed{\frac{x+5}{x^2-7x+10}}\).
Explanation
- Understanding Excluded Values
The excluded values of a rational expression occur where the denominator is equal to zero, as division by zero is undefined in mathematics. To find the excluded values of a given rational expression, we set its denominator to zero and solve for the variable \(x\). In this problem, we are looking for a rational expression that has excluded values of \(x=2\) and \(x=5\). This means the denominator of the correct expression must be zero when \(x=2\) and when \(x=5\). A quadratic polynomial that is zero at \(x=2\) and \(x=5\) must have the form \(k(x-2)(x-5)\), where \(k\) is a non-zero constant. Expanding this, we get \(k(x^2 - 7x + 10)\). Therefore, we need to identify the option whose denominator can be factored into \((x-2)(x-5)\).
- Evaluating the Options
Now we examine each of the provided options:
- \(\frac{x^2-7x+10}{x+5}\): The denominator is \(x+5\). Setting \(x+5=0\) gives \(x=-5\). This does not match our requirements.
- \(\frac{x^2+7x+10}{x+5}\): The denominator is \(x+5\). Setting \(x+5=0\) gives \(x=-5\). This does not match our requirements.
- \(\frac{x+5}{x^2+7x+10}\): The denominator is \(x^2+7x+10\). We can factor this as \((x+2)(x+5)=0\), which gives excluded values \(x=-2\) and \(x=-5\). This does not match our requirements.
- \(\frac{x+5}{x^2-7x+10}\): The denominator is \(x^2-7x+10\). We can factor this as \((x-2)(x-5)=0\). Solving this equation, we get \(x=2\) and \(x=5\). These are the excluded values we were looking for.
- Conclusion
By comparing the findings from the previous step, we see that the expression \(\frac{x+5}{x^2-7x+10}\) is the only one whose denominator vanishes exactly at \(x=2\) and \(x=5\). Thus, it satisfies the condition of having excluded values at \(2\) and \(5\). The correct choice is \(\frac{x+5}{x^2-7x+10}\).
Examples
In real-world applications, such as calculating the average speed of a vehicle where the formula is \(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\), we must ensure that the time is never zero. If the time is expressed as a function, say \(T(x) = x^2 - 7x + 10\), we would need to know which values of \(x\) result in \(T(x) = 0\), as these would represent physical moments where our speed calculation is undefined. Identifying these 'excluded values' is a vital step in ensuring that mathematical models remain valid and meaningful within their defined domain.