To solve the problem, we are given $x = 3 + \sqrt{8}$ and we need to find the value of $x^2 + \frac{1}{x^2}$.

Step-by-Step Solution:

1. Find $x^2$:

   We have: $x = 3 + \sqrt{8}$

   Squaring both sides: $x^2 = (3 + \sqrt{8})^2$

   Expand the square: $x^2 = 3^2 + 2 \cdot 3 \cdot \sqrt{8} + (\sqrt{8})^2$

   Calculate each term: $x^2 = 9 + 6\sqrt{8} + 8$

   Combine like terms: $x^2 = 17 + 6\sqrt{8}$

2. Find $\frac{1}{x}$:

   To find $\frac{1}{x}$, notice that: $x = 3 + \sqrt{8}$

   Then, the conjugate is $3 - \sqrt{8}$. Multiply and divide by the conjugate: $\frac{1}{x} = \frac{1}{3 + \sqrt{8}} \times \frac{3 - \sqrt{8}}{3 - \sqrt{8}}$

   The denominator simplifies to: $(3 + \sqrt{8})(3 - \sqrt{8}) = 3^2 - (\sqrt{8})^2 = 9 - 8 = 1$

   So: $\frac{1}{x} = 3 - \sqrt{8}$

   Now, square $\frac{1}{x}$: $\left( \frac{1}{x} \right)^2 = (3 - \sqrt{8})^2$

   Expand the square: $\left( \frac{1}{x} \right)^2 = 3^2 - 2 \cdot 3 \cdot \sqrt{8} + (\sqrt{8})^2$

   Calculate each term: $\left( \frac{1}{x} \right)^2 = 9 - 6\sqrt{8} + 8$

   Combine like terms: $\frac{1}{x^2} = 17 - 6\sqrt{8}$

3. Find $x^2 + \frac{1}{x^2}$:

   Now, add $x^2$ and $\frac{1}{x^2}$: $x^2 + \frac{1}{x^2} = (17 + 6\sqrt{8}) + (17 - 6\sqrt{8})$

   The terms with $\sqrt{8}$ cancel out: $x^2 + \frac{1}{x^2} = 17 + 17 = 34$

Final Answer:

$x^2 + \frac{1}{x^2} = 34$

Would you like more details on any step, or do you have any questions?

Further Related Questions:

1. How do you simplify expressions involving radicals like $\sqrt{8}$?
2. What is the conjugate of a binomial, and why is it used in simplifying fractions?
3. Can you explain the process of rationalizing the denominator?
4. How would this problem change if $x = 3 - \sqrt{8}$?
5. What are the properties of exponents that are helpful in problems like this?

Tip:

When dealing with expressions involving square roots, always check if you can use conjugates to simplify calculations.