Question

x = 2+ root 3 find x^2 + 1 upon x ^2

Answer 1

ANSWER
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$x = 2 + \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \\ \\ (rationalizing \: the \: denominator \: ) \\ \\ = > \frac{1(2 - \sqrt{3}) }{(2 + \sqrt{3})(2 - \sqrt{3} ) } \\ \\ = > \frac{2 - \sqrt{3} }{( {2}^{2} ) - ( \sqrt{3}^{2} ) } \\ \\ = > \frac{2 - \sqrt{3} }{4 - 3} = > 2 - \sqrt{3} \\ \\ \\ to \: find \: \\ \\ = > x + \frac{1}{x} \\ \\ = > 2 + \sqrt{3} + 2 - \sqrt{3} \\ \\ = > 4 \\ \\ = > x + \frac{1}{x} = 4 \\ \\ {(x + \frac{1}{x} })^{2} = {4}^{2} \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } + 2x (\frac{1}{x} ) = 16 \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 16 \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = 16 - 2 \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = 14 \: \: \: \: \: \: ans$

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#BE BRAINLY

Answer 2

Here is your solution

Given :-

x=2+√3

Now

$\frac{1}{x} = \frac{1}{2 +\sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{2 {}^{2} - ( \sqrt{3}) {}^{2} } \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3} \\ \frac{1}{x} = 2 - \sqrt{3}$

$x + \frac{1}{x} = 2 - \sqrt{3} + 2 + \sqrt{3} \\ x + \frac{1}{x} = 4 \\ Both \: sides \: squaring. \: \\ (x + \frac{1}{x} ) {}^{2} = 4 {}^{2} \\ x {}^{2} + \frac{1}{x {}^{2} } + 2 = 16 \\ x {}^{2} + \frac{1}{x {}^{2} } = 16 - 2 \\ x {}^{2} + \frac{1}{x {}^{2} } = 14$

Hope it helps you