Question

Hi this is for genius

Answer 1

$\textbf{Solution}$

Squaring both sides we have

$\bf{x}^{2} = ( 3 + { \sqrt{8} )}^{2} \\ \\ {x}^{2} = 9 + 8 + 6 \sqrt{8} \\ \\ {x}^{2} = 17 + 12 \sqrt{2} \\ \\ Then \: \: \frac{1}{ {x}^{2} } = 17 - 12 \sqrt{2} \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 17 + 12\sqrt{2} + 17 - 12 \sqrt{2} \\ \\ = 34$

Thanks

Have a colossal day ahead

Be Brainly

Answer 2

$<b>$Answer :

$= > x = 3 + \sqrt{8} \\ \\ lets \: find \: the \: value \: of \: \frac{1}{x} \: first.... \\ \\ = > \frac{1}{x} = \frac{1}{3 + \sqrt{8} } \\ \\ = > \frac{1}{3 + \sqrt{8} } \times \frac{3 - \sqrt{8} }{3 - \sqrt{8} } \\ \\ = > \frac{3 - \sqrt{8} }{ {(3)}^{2} - {( \sqrt{8}) }^{2} } \\ \\ = > \frac{3 - \sqrt{8} }{9 - 8} \\ \\ = > \frac{3 - \sqrt{8} }{1} \\ \\ = > 3 - \sqrt{8}$

Now, let's get the value of...

$= > {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ put \: x = 3 + \sqrt{8} \: \\ = > \frac{1}{x} = 3 - \sqrt{8} \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = {(3 + \sqrt{8}) }^{2} + {(3 - \sqrt{8}) }^{2} \\ \\ = > {(3)}^{2} + {( \sqrt{8} )}^{2} + 2 \times 3 \times \sqrt{8} + ( {(3)}^{2} + {( \sqrt{8} )}^{2} - 2 \times 3 \times \sqrt{8} ) \\ \\ = > 9 + 8 + 6 \sqrt{8} + (9 + 8 - 6 \sqrt{8} ) \\ \\ = > 17 + 6 \sqrt{8} + (17 - 6 \sqrt{8} ) \\ \\ = > 17 + 6 \sqrt{8} + 17 - 6 \sqrt{8 } \\ \\ = > 17 + 17 \\ \\ = > 34$

$\color{pink}\underline\textbf{Hence, 34 is your required answer. }$

Tysm for the question!