Question

please give me answer. I will mark brainliest for 1st​

Answer 1

given:

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

To find :-

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

$\sf (ii) x^{3} + \frac{1}{x^{3} }$

Solution :-

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

Therefore,

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

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

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

$➜\sf \frac{1}{x} = {2 - \sqrt{3} }$

now,

$\sf</u><u> </u><u>x</u><u> + \frac{1}{x}$

$</u><u>➜ \sf 2 + \cancel{\sqrt{3}}+ 2 - \cancel {\sqrt{3}}</u><u>$

$\sf➜2 + 2$

$\sf➜4$

_________________________

$\sf( i)on \: squaring \: both \: sides, we \: have</u><u>:</u><u>$

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

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

$</u><u>\</u><u>boxed</u><u>{</u><u>\sf➜ {x}^{2} + \frac{1}{x^{2} }</u><u> =14</u><u>}</u><u>$

_________________________

$\sf(ii) On \: cubing \: both \: sides \: , we \: have :$

$➜ \sf (x + \frac{1}{x} )^{3} = (4)^{3}$

$➜ \sf x^{3} + \frac{1}{x^{3}} + 3(x + \frac{1}{x} ) = 64$

$➜ \sf x^{3} + \frac{1}{x^{3}} +3 \times 4 = 64$

$➜ \sf x^{3} + \frac{1}{x^{3}} +12 = 64$

$➜ \sf x^{3} + \frac{1}{x^{3}} = 64 - 12$

$\boxed{\sf x^{3} + \frac{1}{x^{3}} = 52}$

$\red {\sf Therefore</u><u>,</u><u>value \: of \: x^{2} + \frac{1}{x^{2}} = 14 \: and \: x^{3} + \frac{1}{x^{3}} = 52.}$