\( x=3+2 \sqrt{2} \), find the value of \( \left(x^{2}+\frac{1}{x^{2}}\right) \)

To find the value of \( \left(x^{2} + \frac{1}{x^{2}}\right) \) where \( x = 3 + 2\sqrt{2} \), we can follow these steps: First, calculate \( x^2 \): \[ x^2 = (3 + 2\sqrt{2})^2 = 3^2 + 2 \cdot 3 \cdot 2\sqrt{2} + (2\sqrt{2})^2 \] \[ = 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2} \] Next, find \( \frac{1}{x} \): To find \( \frac{1}{x} \), we can multiply the numerator and denominator by the conjugate: \[ \frac{1}{x} = \frac{1}{3 + 2\sqrt{2}} \cdot \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} \] Calculating the denominator: \[ (3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 9 - 8 = 1 \] So, \( \frac{1}{x} = 3 - 2\sqrt{2} \). Now we find \( \frac{1}{x^2} \): \[ \frac{1}{x^2} = (3 - 2\sqrt{2})^2 = 3^2 - 2 \cdot 3 \cdot 2\sqrt{2} + (2\sqrt{2})^2 \] \[ = 9 - 12\sqrt{2} + 8 = 17 - 12\sqrt{2} \] Now, we can calculate \( x^2 + \frac{1}{x^2} \): \[ x^2 + \frac{1}{x^2} = (17 + 12\sqrt{2}) + (17 - 12\sqrt{2}) = 34 \] Thus, the value of \( \left(x^{2} + \frac{1}{x^{2}}\right) \) is \[ \boxed{34} \]