Question

if x is equal to 1 + root 2 find the value of x square + 1 upon x square
$if x = 1 + \sqrt{2 \: find \: the \: value of {x}^{2} + \frac{1}{ {x}^{2} }$

Answer 1

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$x = 1 + \sqrt{2} \\ \frac{1}{x} = \frac{1 }{1 + \sqrt{2} } = \frac{( \sqrt{2} - 1)}{( \sqrt{2} - 1)( \sqrt{2} + 1) } \\ = \frac{ \sqrt{2} - 1 }{( { \sqrt{2}) }^{2} - {1}^{2} } = \frac{ \sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1 \\ \\ so.. \: \: \: {x}^{2} + \frac{1}{ {x}^{2} } \\ = (x + \frac{1}{x} )^{2} - 2x. \frac{1}{x} \\ = (1 + \sqrt{2} + \sqrt{2} - 1)^{2} - 2 \\ = {(2 \sqrt{2} )}^{2} - 2 = 8 - 2 \\ = 6$
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