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:

\[ x \in (-\infty, -6) \cup (6, \infty) \]