X2-36>0

To solve the inequality \( x^2 - 36 > 0 \), we first factor the left side. The expression \( x^2 - 36 \) can be factored as follows: \[ x^2 - 36 = (x - 6)(x + 6) \] So the inequality becomes: \[ (x - 6)(x + 6) > 0 \] Next, we need to determine where the product \( (x - 6)(x + 6) \) is greater than zero. We can find the critical points by setting each factor to zero: \[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \] \[ x + 6 = 0 \quad \Rightarrow \quad x = -6 \] These critical points divide the number line into three intervals: 1. \( (-\infty, -6) \) 2. \( (-6, 6) \) 3. \( (6, \infty) \) Now we will test each interval to see where the product is positive. 1. **Interval \( (-\infty, -6) \)**: - Choose \( x = -7 \): \[ (-7 - 6)(-7 + 6) = (-13)(-1) = 13 > 0 \] - So, \( (x - 6)(x + 6) > 0 \) in this interval. 2. **Interval \( (-6, 6) \)**: - Choose \( x = 0 \): \[ (0 - 6)(0 + 6) = (-6)(6) = -36 < 0 \] - So, \( (x - 6)(x + 6) < 0 \) in this interval. 3. **Interval \( (6, \infty) \)**: - Choose \( x = 7 \): \[ (7 - 6)(7 + 6) = (1)(13) = 13 > 0 \] - So, \( (x - 6)(x + 6) > 0 \) in this interval. Now we can summarize the results: - \( (x - 6)(x + 6) > 0 \) for \( x \in (-\infty, -6) \) and \( x \in (6, \infty) \). Thus, the solution to the inequality \( x^2 - 36 > 0 \) is: \[ x \in (-\infty, -6) \cup (6, \infty) \]