Question

answer please if u want test question before exam​

Answer 1

$\huge\underline\mathrm{Given:-}$

$\: \: \: \: x = 2 + \sqrt{3}$

$\frac{1}{x} = \frac{1}{2 + \sqrt{3} }$

$\: \: \: = 1 \times \frac{(2 - \sqrt{3)} }{(2 + \sqrt{3})(2 - \sqrt{3}) }$

$\: \: = \frac{ (2 - \sqrt{3})}{ {(2}^{2} - { (\sqrt{3}) }^{2} ) }$

$\: \: = \frac{(2 - \sqrt{3)} }{4 - 3}$

$\: \: = 2 - \sqrt{3}$

$\huge\underline\mathrm{Solution:-}$

${x}^{2} = (2 + \sqrt{3} )$

$\: \: = {(2)}^{2} + {( \sqrt{3)} }^{2} + 2 \times 2 \times \sqrt{3}$

$\: \: = 4 + 3 + 4 \sqrt{3}$

$\: \: 7 + 4 \sqrt{3}$

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

$\: \: = {(2)}^{2} + { (\sqrt{3}) }^{2} - 2 \times 2 \times \sqrt{3}$

$\: \: = 4 + 3 - 4 \sqrt{3}$

$\: \: = 7 - 4 \sqrt{3}$

$\mathrm{Now,}$

${ x }^{2} + \frac{1}{ {x}^{2} } = (7 + 4 \sqrt{3} ) + (7 - 4 \sqrt{3} )$

$\: \: = 7 + 4 \sqrt{3} + 7 - 4 \sqrt{3}$

$\: \: = 7 + 7 + 4 \sqrt{3} - 4 \sqrt{3}$

$\: \: = 14$

hope it's helpful..

Answer 2

Answer:

$\huge\sf\purple{Hope It Helps}$