We are tasked with solving the equation $x^2 + \frac{1}{x^2} = 81$ and finding the value of $x + \frac{1}{x}$. Let's solve step by step:

Step 1: Let $y = x + \frac{1}{x}$.

Squaring both sides gives: $y^2 = \left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2.$

From the given equation, $x^2 + \frac{1}{x^2} = 81$. Substituting this into the equation above, we get: $y^2 = 81 + 2 = 83.$

Step 2: Solve for $y$.

$y = \pm \sqrt{83}.$

Thus, the value of $x + \frac{1}{x}$ is $\pm \sqrt{83}$.

Would you like a detailed verification or clarification of any step? Here are some follow-up questions:

1. How can we verify $y^2 = 83$ from the original equation?
2. Can this problem be solved using different substitutions?
3. How does the sign $\pm$ depend on the value of $x$?
4. What does this solution imply about the relationship between $x$ and $\frac{1}{x}$?
5. Are there any alternative ways to simplify $x^2 + \frac{1}{x^2}$?

Tip: Squaring and substituting is a common technique when solving equations involving fractions and powers of variables.