Math, asked by sandeeppabbi, 1 year ago

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

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Answered by Saksham0505
3

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Answered by Anonymous
2

 \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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