Question

Hello ! ❤❤



If x = 2 + √3

Find :

(i) X + 1 /x

(ii) X² + 1/x²

(iii) X³ + 1/ x³



thanks ! ☺☺☺

Answer 1

Hi there !

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Given :

x = 2 + √3

To Find :

(i) x + 1 / x

(ii) x² + 1 / x²

(iii) x³ + 1 / x³

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Question I :

x = 2 + √3

1 / x = 1 / 2 + √3

1 / x = 1 / 2 + √3 × 2 - √3 / 2 - √3

1 / x = 2 - √3 / (2)² - (√3)²

1 / x = 2 - √3 / 4 - 3

1 / x = 2 - √3

Now,

x + 1 / x = 2 + √3 + 2 - √3

x + 1 / x = 2 + 2

x + 1 / x = 4  [Ans]

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Question II :

x + 1/x = 4

On squaring both sides ..

( x + 1/x )² = (4)²

x² + 1 / x² + 2 = 16

x² + 1 / x² = 16 - 2

x² + 1 / x² = 14  [Ans]

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Question III :

x + 1 / x = 4

On cubing both sides ..

( x + 1/x )³ = (4)³

x³ + 1/x³ + 3 ( x + 1/x) = 64

x³ + 1/x³ + 3 (4) = 64

x³ + 1/x³ + 12  = 64

x³ + 1/x³ = 64 - 12

x³ + 1/x³ = 52  [Ans]

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Thanks for the question !

Answer 2

$\bf{\huge{\underline{ANSWER}}}$

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$<b>GIVEN : x = 2 + √3 _________________________ SOLUTION :<b>$

$\frac{1}{x} = \frac{1}{2 + \sqrt{3} }$
Rationalising the denominator , we get



$\frac{(1) \: (2 - \sqrt{3}) }{(2 + \sqrt{3} ) \: (2 - \sqrt{3} )} \\ \\ = > \: \frac{(2 - \sqrt{3}) }{ {2}^{2} - { \sqrt{3} }^{2} } \\ \\ = > \frac{2 - \sqrt{3} }{4 - 3} \: \: \: \: = > 2 - \sqrt{3}$
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TO FIND :

( i )
$x + \frac{1}{x} \\ \\ = > 2 + \sqrt{3} + 2 - \sqrt{3} \\ \\ = > 4 \: \: \: \: \: \\ \\ = > \: x + \frac{1}{x} = 4 \: \: \: \: \: \: ans$

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

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( iii )
${x}^{3} + \frac{1}{ {x}^{3} } \\ \\ = > x + \frac{1}{x} = 4 \\ \\ = > {(x + \frac{1}{x} })^{3} = ( {4)}^{3} \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } + 3(x)( \frac{1}{x} )(x + \frac{1}{x} ) = 64 \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } + 3 \: (4) = 64 \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } = 64 - 12 \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } = 52 \: \: \: \: \: ans$

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Hope it helps....

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