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If x = 3 + 2$$\sqrt{\ 2}$$ what will be the value of $$X^2 + \left(\frac{1}{X^2}\right)$$?
35
32
36
34
X=3+2$$\sqrt{\ 2}$$
$$\frac{1}{x}=3-2\sqrt{\ 2}$$
$$x^2+\frac{1}{x^2}=\left(x\ +\frac{1}{x}\right)^2-2$$ = $$6^2-2=34$$
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