Question

$x = 2 + \sqrt{3} \: find \\ \frac{1}{x} \\ x + \frac{1}{x} \\ x - \frac{1}{x} \\ {x}^{2} + \frac{1}{x2}$
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Answer 1

Answer:

$x = 2 + \sqrt{3} \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \\ x + \frac{1}{x}\\ = 2 + \sqrt{3} + \frac{1}{2 + \sqrt{3} } \\ = \frac{ {(2 + \sqrt{3}) }^{2} + 1}{2 + \sqrt{3} } \\ = \frac{4 + 3 + 4 \sqrt{3} + 1 }{2 + \sqrt{3} } \\ = \frac{8 + 4\sqrt{3} }{2 + \sqrt{3} } \\ = \frac{4(2 + \sqrt{3} )}{(2 + \sqrt{3}) } \\ = 4 \\ {x}^{2} + \frac{1}{ {x}^{2} } \\ = {(2 + \sqrt{3} )}^{2} + \frac{1}{ {(2 + \sqrt{3} )}^{2} } \\ = 4 + 3 + 4 \sqrt{3} + \frac{1}{4 + 3 + 4 \sqrt{3} } \\ = 7 + 4 \sqrt{3} + \frac{1}{7 + 4 \sqrt{3} } \\ = \frac{ {(7 + 4 \sqrt{3} )}^{2} + 1 }{7 + 4 \sqrt{3} } \\ = \frac{49 + 48 + 56 \sqrt{3} + 1}{7 + 4 \sqrt{3} } \\ = \frac{98 + 56 \sqrt{3} }{7 + 4 \sqrt{3} } \\ = \frac{14(7 + 4 \sqrt{3} )}{(7 + 4 \sqrt{3}) } \\ = 14$

Sorry, but I am not getting the answer for x-1/x.