Question

If x = root 3 + root 2 / root 3 - root 2 , find x^4 + 1 / x^4​

Answer 1

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

$- > \frac{( \sqrt{3} + \sqrt{2})( \sqrt{3} + \sqrt{2}) }{( \sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2})}$

$- > \frac{3 + \sqrt{6} + \sqrt{6} + 2 }{ {( \sqrt{3} )}^{2} - {( \sqrt{2}) }^{2} }$

$- > \frac{5 + 2 \sqrt{6} }{3 - 2}$

$- > \frac{5 + 2 \sqrt{6} }{1}$

$- > x = 5 + 2 \sqrt{6} - - > 1$

Now ,

$\frac{1}{x} = \frac{1}{5 + 2 \sqrt{6} } (from \: 1)$

$- > \frac{1 \times 5 - 2 \sqrt{6} }{(5 + 2 \sqrt{6)(5 - 2 \sqrt{6} )} }$

$- > \frac{5 - 2 \sqrt{6} }{ {(5) }^{2} - {(2 \sqrt{6} )}^{2} }$

$- > \frac{5 - 2 \sqrt{6} }{25 - 24}$

$- > \frac{5 - 2 \sqrt{6} }{1}$

$- > \frac{1}{x} = 5 - 2 \sqrt{6} - - > 2$

Now ,

$x + \frac{1}{x} = 5 + 2 \sqrt{6} + 5 - 2 \sqrt{6}$

$- > x + \frac{1}{x} = 10$

Please refer the above attachment fir the further answer..

The last part of the answer is in the above attachement....

I hope it helps u dear friend ^_^♡♡