Question

x is equal to 2 + root 3 then find the value of x square + 1 upon x square

Answer 1

Heya !!!



X = 2 + ✓3



1/X = 1/2+✓3



1/X = 1/2 + ✓3 × ( 2 - ✓3 ) / ( 2-✓3 )



1/X = ( 2 - ✓3 ) / ( 2 + ✓3 ) ( 2 - ✓3 )


1/X = 2-✓3/ (2)² - ( ✓3 )²


1/X = ( 2 - ✓3 )



Therefore,


X + 1/X = ( 2 + ✓3 ) + ( 2-✓3 ) = 4




Squaring both sides,


( X + 1/X)² + (4)²



X² + 1/X² + 2 × X × 1/X = 16




X² + 1/X² = 16-2



X² + 1/X² = 14

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Answer 2

x = 2 + √3

$\frac{1}{x} = \frac{1}{2 + \sqrt{3} }$

By Rationalization,

$\frac{1}{x} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \\ => \frac{1}{x} = \frac{2 - \sqrt{3} }{(2 + \sqrt{3} )(2 - \sqrt{3}) } \\ \\ \\ = > \frac{1 }{x} = \frac{2 - \sqrt{3} }{ {2}^{2} - {( \sqrt{3} )}^{2} } \\ \\ \\ = > \frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3} \\ \\ \\ = > \frac{1}{x} = \frac{2 - \sqrt{3} }{1} \\ \\ \\ = > \frac{1}{x} = 2 - \sqrt{3}$


Hence,


$x + \frac{1}{x} = 2 + \sqrt{3} + 2 - \sqrt{3} \\ \\ = > x + \frac{1}{ x} = 4$



Now, Square on both sides,


${(x + \frac{1}{x}) }^{2} = {4}^{2} \\ \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } + 2(x \times \frac{1}{x} ) = 16 \\ \\ \\ => {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 16 \\ \\ \\ => {x}^{2} + \frac{1}{ {x}^{2} } = 16 - 2 = 14$