Question

Hii friends!!! plzz give me proper answers with proper method....!! plzz dont give me any improper answers..........

if x= √3 +√2 upon √3 - √2, find x^2 +1 upon x^2 ????

Answer 1

Heya Frnd ...........☺

Answer is

$= \boxed{98}$

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HOPE IT WILL HELP U.............✔

Answer 2

HELLO DEAR,

$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ = > x = \frac{( \sqrt{3} + \sqrt{2} )( \sqrt{3} + \sqrt{2} ) }{( \sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2}) } \\ \\ = > x = \frac{ {( \sqrt{3} + \sqrt{2} )}^{2} }{3 - 2} \\ \\ = > x = \frac{3 + 2 + 2 \times \sqrt{3} \times \sqrt{2} }{1} \\ \\ = > x = 5 + 2 \sqrt{6} \\ \\ on \: squaring \: both \: side \\ \\ {x}^{2} = 25 + 24 + 20 \sqrt{6} \\ \\ = > {x}^{2} = 49 + 20 \sqrt{6} .......(1)$


${x}^{2} + \frac{1}{ {x}^{2} } \\ \\ = > 49 + 20 \sqrt{6} + \frac{1}{49 + 20 \sqrt{6} } \times \frac{49 - 20 \sqrt{6} }{49 - 20 \sqrt{6} }......from..(1) \\ \\ = > 49 + 20 \sqrt{6} + \frac{49 - 20 \sqrt{6} }{2401 - 2400} \\ \\ = > 49 + 20 \sqrt{6} + \frac{49 - 20 \sqrt{6} }{1} \\ \\ = > 49 + 20 \sqrt{6} + 49 - 20 \sqrt{6} \\ \\ = > 98$


I HOPE ITS HELP YOU DEAR,
THANKS