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2015-04-23T11:00:19+05:30

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The trick is to use the identityx^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2

In view of using above identity, start by finding the value of x+\frac{1}{x} :
x=2+\sqrt{3}\\~\\\frac{1}{x}=\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{2^2-\sqrt{3}^2}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}

Adding these
x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4

Plug in the earlier identity and get
x^2+\frac{1}{x^2}=4^2-2=14
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