Question

$x = 2 + \sqrt{3} find \: the \: value \: x {}^{2} + \frac{1}{ {x}^{2} }$

Answer 1

x = 2 + √3 

x² = (2 + 3)²

x² = 4 + 3 + 4√3

x²  = 7 + 4√3
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1/x = 1/(2 + √3)

               By Rationalization

1/x = (2 - √3) / [ (2)² - (√3)²]


1/x = (2 - √3)/ (4 - 3)

1/x = (2 - √3)/1 

1/x = 2 - √3 

1/x² = (2 - 3)² 

1/x² = 4 + 3 -4√3

1/x² = 7 - 4√3


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Now,

x²  + 1/x² 

7 + 4√3 + 7 - 4√3 

7 + 7 

14 


i hope this will help you


(-:

Answer 2

Heya !!

Here's your answer..
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Given :- x = 2 + √3

To Find :- x² + 1/x²

Solution :-
$x = 2 + \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3} \\ \\ \frac{1}{x} = 2 - \sqrt{3} \\ \\ x + \frac{1}{ x} = 2 + \sqrt{3} + 2 - \sqrt{3} \\ \\ x + \frac{1}{x} = 4 \\ \\ take \: \: square \: \: on \: \: both \: \: side \\ \\ {(x + \frac{1}{x} )}^{2} = {(4)}^{2} \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2x \times \frac{1}{x} = 16 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 16 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 16 - 2 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 14$

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Hope it helps..
Thanks :)