If \( x=3-2 \sqrt{2} \) then find the value of \( x^{2}+\frac{1}{x^{2}} \)

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