Question

if x is equal to 2 + under root 3 find X raise to power 2 + 1 by X raise to power 2​

Answer 1

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

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

rationalise the denominator

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

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

$\sf{\frac{1}{x} = \frac{2 - \sqrt{3} }{ 4 - 3 } }$

$\sf{\frac{1}{x} = {2 - \sqrt{3} }}$

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

$\sf{x + \frac{1}{x} = 4}$

squaring on both sides

$\sf{{(x + \frac{1}{x} ) }^{2} = {4}^{2} }$

$\sf{{ {x}^{2} + \frac{1}{{x}^{2}} + 2= 16}}$

$\sf{{ {x}^{2} + \frac{1}{{x}^{2}} = 16 - 2}}$

$\fbox{\sf{{ {x}^{2} + \frac{1}{{x}^{2}} = 14}}}$

Answer 2

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