If \( x=2+\sqrt{3} \), find the value of \( x^{2}+\frac{1}{x^{2}} \)
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To find the value of \( x^{2} + \frac{1}{x^{2}} \) where \( x = 2 + \sqrt{3} \), we start by calculating \( x^2 \). \[ x^2 = (2 + \sqrt{3})^2 = 2^2 + 2 \cdot 2 \cdot \sqrt{3} + (\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} \] Next, we need \( \frac{1}{x} \): \[ \frac{1}{x} = \frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3} \] Now, we calculate \( \frac{1}{x^2} \): \[ \frac{1}{x^2} = (2 - \sqrt{3})^2 = 2^2 - 2 \cdot 2 \cdot \sqrt{3} + (\sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3} \] Now we can find \( x^2 + \frac{1}{x^2} \): \[ x^2 + \frac{1}{x^2} = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 7 + 4\sqrt{3} + 7 - 4\sqrt{3} = 14 \] Thus, the value of \( x^{2} + \frac{1}{x^{2}} \) is \(\boxed{14}\).