Question

plz solve this guy's

Answer 1

Hey There!!

Here we have:

$x=3+\sqrt{8} \\ \\ \\ \implies \frac{1}{x} = \frac{1}{3+\sqrt{8}} \\ \\ \\ \implies \frac{1}{x} = \frac{1}{3+\sqrt{8}} \times \frac{3-\sqrt{8}}{3-\sqrt{8}} \\ \\ \\ \implies \frac{1}{x} = \frac{3-\sqrt{8}}{9-8} \\ \\ \\ \implies \frac{1}{x} = 3-\sqrt{8}$


$\implies x+\frac{1}{x} = (3+\sqrt{8}) + (3-\sqrt{8}) \\ \\ \\ \implies x+\frac{1}{x} = 6$



Now,

$x^2+\frac{1}{x^2} \\ \\ \\ = x^2 + \frac{1}{x^2} +2 - 2 \\ \\ \\ = x^2 + 2(x) \left(\frac{1}{x} \right) + \frac{1}{x^2} - 2 \\ \\ \\ = \left(x+\frac{1}{x} \right)^2 - 2 \\ \\ \\ = 6^2 - 2 \\ \\ \\ = 36 - 2 \\ \\ \\ = 34 \\ \\ \\ \\ \\ \implies \boxed{x^2+\frac{1}{x^2}=34}$


Hope it helps
Purva
Brainly Community

Answer 2

x = 3 + √8

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

By Rationalization,

$\frac{1}{x} = \frac{1}{3 + \sqrt{8} } \times \frac{3 - \sqrt{8} }{ 3 - \sqrt{8} } \\ \\ \\ \frac{1}{x} = \frac{3 - \sqrt{8} }{(3 + \sqrt{8} )(3 - \sqrt{8} )}$


××××××××××××××
We know,
(a + b)(a - b) = a² - b²
××××××××××××××××

Applying formula on denominator,


$\frac{1}{x} = \frac{ 3 - \sqrt{8} }{ {3}^{2} - {( \sqrt{8}) }^{2} } \\ \\ \\ \frac{1}{x} = \frac{3 - \sqrt{8} }{9 - 8} \\ \\ \\ \frac{1}{x} = \frac{3 - \sqrt{8} }{1} \\ \\ \\ \frac{1}{x} = 3 - \sqrt{8}$



Now,


${x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ => (3 + \sqrt{8} )^{2} + {(3 - \sqrt{8}) }^{2}$

×××××××××××××××××
We know,
(a + b)² = a² + b² + 2ab
(a - b)² = a² + b² - 2ab
××××××××××××××××××××

Applying formula,


=> (3 + √8)² + (3 - √8)²

=> (3)² + (√8)² + 6√8 + (3)² - (√8)² - 6√8

=> 9 + 8 + 6√8 + 9 + 8 - 6√8

=> 9 + 8 + 9 + 8

=> 17 + 17

=> 34



I hope this will help you

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