Question

please answer this 17 th que

Answer 1

I hope it will help you

Answer 2

$Given : x = 3 + \sqrt{8}$

$= \ \textgreater \ \frac{1}{x} = \frac{1}{3 + \sqrt{8} } * \frac{3 - \sqrt{8} }{3 - \sqrt{8} }$

$= \ \textgreater \ \frac{3 - \sqrt{8} }{(3)^2 - ( \sqrt{8})^2 }$

$= \ \textgreater \ \frac{3 - \sqrt{8} }{9 - 8}$

$= \ \textgreater \ 3 - \sqrt{8}$

$= \ \textgreater \ x + \frac{1}{x} = 3 + \sqrt{8} + 3 - \sqrt{8}$

$= \ \textgreater \ x + \frac{1}{x} = 6$

On squaring both sides, we get

$= \ \textgreater \ x^2 + \frac{1}{x^2} + 2 = 36$

$= \ \textgreater \ x^2 + \frac{1}{x^2} = 34$

On squaring both sides, we get

$= \ \textgreater \ (x^ 4 + \frac{1}{x^4} ) + 2 = 1156$

$= \ \textgreater \ x^4 + \frac{1}{x^4} = 1156 - 2$

$= \ \textgreater \ x^4 + \frac{1}{x^4} = 1154$




Hope this helps!