Question

If x = 5+2√6 / 5−2√6 , find the value of 
(1)(x + 1/x)^2 
(2) (x − 1/x)^2

Answer 1

Here's your answer!!

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It's given that,

$= > x = \frac{5 + 2 \sqrt{6} }{5 - 2 \sqrt{6} }$

We will rationalise it ,

$= > \frac{5 + 2 \sqrt{6} }{5 - 2 \sqrt{6} } \times \frac{5 + 2 \sqrt{6} }{5 + 2 \sqrt{6} }$

$= > \frac{ {(5 + 2 \sqrt{6}) }^{2} }{ {(5)}^{2} - {(2 \sqrt{6}) }^{2} }$

$= > \frac{25 + 24 + 20 \sqrt{6} }{25 - 24}$

$= > {x} = 49 + 20 \sqrt{6}.....(a)$

Now,

$= > \frac{1}{x} = \frac{1}{49 + 20 \sqrt{49} }$

By rationalizing it ,we get :-

$= > \frac{1}{49 + 20 \sqrt{6} } \times \frac{49 - 20 \sqrt{6} }{49 - 20 \sqrt{6} }$

$= > \frac{49 - 20 \sqrt{6} }{ {(49)}^{2} - {(20 \sqrt{6}) }^{2} }$

$= > \frac{49 - 20 \sqrt{6} }{2401 - 2400}$

$= > \frac{1}{x} = 49 - 20 \sqrt{6}....(b)$

We will find the value of ,

$= > (x + \frac{1}{x} )$

From a and b

$= > (x + \frac{1}{x}) = 49 + 20 \sqrt{6} + 49 - 20 \sqrt{6}$

$- 20 \sqrt{6} and \: 20 \sqrt{6} get \: cancelled$

$= > (x + \frac{1}{x} ) = 98$

By squaring both side ,we get :-

$= > {(x + \frac{1}{x}) }^{2} = {(98)}^{2}$

$= > {(x + \frac{1}{x} )}^{2} = 9604......(1 \: answer)$

Now,

From a and b ,

$(x - \frac{1}{x}) = (49 + 20 \sqrt{6}) - (49 - 20 \sqrt{6} )$

$(x - \frac{1}{x}) = 49 + 20 \sqrt{6} - 49 + 20 \sqrt{6}$

$49 \: and \: - 49 \: get \: cancelled$

$(x - \frac{1}{x}) = 40 \sqrt{6}$

By squaring both side ,we get :-

${(x - \frac{1}{x} )}^{2} = {(40 \sqrt{6}) }^{2}$

${(x - \frac{1}{x} )}^{2} = 9600......(2 \: answer)$

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Hope it helps you!! :)

Answer 2

$<b > Ello!¡$

• Check the attachment please.

Hope it helps!! :)

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