Yes, exactly. Let's go through the proper simplification process step-by-step.

Given expression:
\[
\frac{\sqrt{n^2 - 36}}{n + 6}
\]

### Simplification:

1. Recognize that \(n^2 - 36\) is a difference of squares:
   \[
   n^2 - 36 = (n - 6)(n + 6)
   \]

2. Substitute this back into the original expression:
   \[
   \frac{\sqrt{(n - 6)(n + 6)}}{n + 6}
   \]

3. The square root of a product can be written as the product of the square roots:
   \[
   \frac{\sqrt{(n - 6)(n + 6)}}{n + 6} = \frac{|n - 6|}{n + 6}
   \]
   Here, \(|n - 6|\) is the absolute value of \(n - 6\).

### Why Use Absolute Value:

The square root function always returns a non-negative value, hence \(\sqrt{(n - 6)^2} = |n - 6|\). This ensures the result is always non-negative.

### Final Expression:

The simplified expression is:
\[
\frac{|n - 6|}{n + 6}
\]

This fraction takes different forms depending on the value of \(n\):
- For \(n \geq 6\), \(|n - 6| = n - 6\), so the expression simplifies to:
  \[
  \frac{n - 6}{n + 6}
  \]
- For \(n < 6\), \(|n - 6| = 6 - n\), so the expression simplifies to:
  \[
  \frac{6 - n}{n + 6}
  \]

### Summary:

The original expression can be simplified to \(\frac{|n - 6|}{n + 6}\). This correctly reflects the absolute value required when simplifying the square root of a squared term.

Would you like more details or have any questions? Here are some questions you might ask next:

1. How do you handle the absolute value in different cases of \(n\)?
2. What are the steps to verify if an algebraic simplification is correct?
3. Can you provide examples of simplifying other algebraic expressions involving square roots?
4. What are some common errors to avoid in algebraic simplifications?
5. How do you factor and simplify expressions with higher powers?

**Tip:** Always remember to include the absolute value when taking the square root of a squared expression to ensure the result is non-negative.

Learn how to simplify the expression sqrt(n^2 - 36) / (n + 6) step-by-step with absolute value considerations. This problem involves algebraic simplification using the difference of squares formula, suitable for high school students.

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