Okay, let's tackle this problem step by step. The question is asking: "For how many values of x is the expression (x - 5)/(x² - 9) not defined?" Hmm, right. So, first, I need to remember when a rational expression like this isn't defined. I think it's when the denominator is zero because you can't divide by zero. So, the expression is undefined whenever the denominator equals zero. Therefore, I need to find all the x values that make x² - 9 equal to zero and count them. 

Wait, but before I jump into solving the equation, let me make sure I understand the problem correctly. The expression is a fraction, so if the denominator is zero, the entire expression is undefined, regardless of the numerator. So, even if the numerator is also zero, the expression is still undefined. For example, 0/0 is undefined. So, in this case, we just need to find the x values that make the denominator zero. 

So, the denominator is x² - 9. Let me write that down: x² - 9 = 0. To solve this equation, I can factor the left side. Since it's a difference of squares, right? Because x² is a square and 9 is 3². The difference of squares formula is a² - b² = (a - b)(a + b). So, applying that here, x² - 9 becomes (x - 3)(x + 3). Therefore, the equation factors to (x - 3)(x + 3) = 0. 

Now, to find the solutions, I set each factor equal to zero. So, x - 3 = 0 gives x = 3, and x + 3 = 0 gives x = -3. Therefore, the denominator is zero when x is 3 or -3. Hence, the expression is undefined at these two points. 

But wait, let me double-check. The original denominator is x² - 9. Plugging in x = 3: 3² - 9 = 9 - 9 = 0. Correct. Plugging in x = -3: (-3)² - 9 = 9 - 9 = 0. Also correct. So, both x = 3 and x = -3 make the denominator zero, so the expression is undefined at those points. 

Therefore, there are two values of x where the expression is not defined. 

But hold on, let me make sure there aren't any other values that could make the denominator zero. The denominator is a quadratic equation, so it can have at most two real roots. Since we already found both roots, there are exactly two values. Therefore, the answer should be 2. 

Wait, but let me think again. The numerator is x - 5. If the numerator and denominator both equal zero at some x, would that affect the undefined nature? For example, if x - 5 = 0 and x² - 9 = 0 at the same x, then the expression would be 0/0, which is indeterminate. However, in this case, solving x - 5 = 0 gives x = 5. But x² - 9 = 0 gives x = 3 or x = -3. So, x = 5 is not a solution to the denominator equation. Therefore, the numerator and denominator don't share any common roots. Hence, at x = 3 and x = -3, the denominator is zero, but the numerator is not. Therefore, the expression is undefined at both of these points, and there's no overlap where both numerator and denominator are zero. 

Therefore, the number of values of x where the expression is not defined is exactly 2. 

But just to be thorough, let me verify once more. Let's substitute x = 3 into the denominator: 3² - 9 = 0. Yes. Substitute x = -3: (-3)² - 9 = 0. Also yes. So, both are valid solutions. There are no other solutions because the quadratic equation x² - 9 = 0 has exactly two real roots, 3 and -3. 

Therefore, the answer is 2. 

Wait, but maybe I should check if there's any other way the expression could be undefined. For example, sometimes expressions can have other undefined behaviors, like taking the square root of a negative number or something. But in this case, the expression is a fraction with a polynomial in the denominator. The only undefined points are where the denominator is zero. There are no square roots or logarithms here, so that's not an issue. 

Therefore, I can confidently say that there are two values of x where the expression is undefined. 

Final check: The problem asks for "how many values of x", so the answer is the count of these values. Since there are two distinct real roots, the answer is 2. 

Therefore, the final answer is 2.
- The expression is undefined when the denominator is zero.
- The denominator \( x^2 - 9 \) factors to \( (x - 3)(x + 3) \).
- Setting each factor equal to zero gives the solutions \( x = 3 \) and \( x = -3 \).
- There are two distinct values of \( x \) that make the expression undefined.
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