Question

x=3+root 8 find the value of x square +1/x square

Answer 1

$\pink{ \underline{ \underline {\purple{\huge{ \mathfrak{\star{heya \: buddy\star}} }}}}}$

$x = 3 + \sqrt{8}$
$\frac{1}{x} = \frac{1}{3 + \sqrt{8} } \\ = \frac{1}{3 + \sqrt{8} } \times \frac{3 - \sqrt{8} }{3 - \sqrt{8} } \\ = \frac{3 - \sqrt{8} }{9 - 8} \\ = \frac{3 - \sqrt{8} }{1} \\ = 3 - \sqrt{8}$
${x}^{2} + {( \frac{1}{x}) }^{2} \\ = {(3 + \sqrt{8}) }^{2} + {(3 - \sqrt{8}) }^{2} \\ = 9 + 8 + 6 \sqrt{8} + 9 + 8 - 6 \sqrt{8} \\ = 34$
Thus, 34 will be the final answer!

♥️${Be } ^{Brainly }$♥️

Answer 2

Hey there!

___________________

Given :

x = 3 + √8

To find :

$x {}^{2} + \frac{1}{x {}^{2} }$

Solution:

$x = 3 + \sqrt{8} \\ \\ \frac{1}{x} = \frac{1}{3 + \sqrt{8} } \times \frac{3 - \sqrt{8} }{3 - \sqrt{8} } \\ \\ \frac{1}{x} = \frac{3 - \sqrt{8} }{(3) {}^{2} - ( \sqrt{8} ) {}^{2} } \\ \\ \frac{1}{x} = \frac{3 - \sqrt{8} }{9 - 8} \\ \\ \frac{1}{x} = 3 - \sqrt{8} \\ \\ \bold{now..} \\ \\ x + \frac{1}{x} = 3 + \cancel{ \sqrt{8}} + 3 - \cancel{ \sqrt{8}} \\ \\ x + \frac{1}{x} =3 + 3 \\ \\ x + \frac{1}{x} =6\\ \\ \bold{on \: squaring \: both \: sides..} \\ \\ (x + \frac{1}{x} ) {}^{2} = (6) {}^{2} \\ \\ x {}^{2} + \frac{1}{x {}^{2} } + 2 \times \cancel x \times \frac{1}{ \cancel x} = 36 \\ \\ x {}^{2} + \frac{1}{x {}^{2} } + 2 = 36 \\ \\ x {}^{2} + \frac{1}{x {}^{2} } = 36 - 2 \\ \\ \boxed{ \bold{ x {}^{2} + \frac{1}{x {}^{2} } = 34}}$.

___________________________

Thanks for the question!