Math, asked by sandeeppabbi, 11 months ago

if x= 1/2-√3 then find the value of (x-1/x)

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Answered by Saksham0505

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Answered by Anonymous

$\sf \underline{given } : \: \: x = \frac{1}{2 - \sqrt{3} }$

$\sf \underline{to \: find } : \: \: x - \frac{1}{x}$

$\sf x = \frac{1}{2 - \sqrt{3} }$

rationalise the denominator.

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

$\sf x = \frac{2 + \sqrt{3} }{ {2}^{2} - {(\sqrt{3})}^{2} }$

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

$\sf x = {2 + \sqrt{3} }$

now,

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

rationalise the denominator.

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

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

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

$\sf \frac{1}{x} = {2 - \sqrt{3} }$

Hence,

$\sf x - \frac{1}{x} = 2 + \sqrt{3} - (2 - \sqrt{3} )$

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

$\fbox{\sf x - \frac{1}{x} = 2 \sqrt{3}}$

sandeeppabbi: okay

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