Question

x=1/2-√3,then find the value of [x-1/x]​

Answer 1

GIVEN :-

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

TO FIND :-

$x - \frac{1}{x}$

SOLUTION :-

First we shall find the value of x by rationalising the denominator ;

$\frac{1}{2 - \sqrt{3} } \times 1 \\ \\ \\ \\ = \frac{1}{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } \\ \\ \\ \\ = \frac{2 + \sqrt{3} }{ {2}^{2} -{ \sqrt{3} }^{2} } \\ \\ \\ \\ = \frac{2 + \sqrt{3} }{4 - 3} \\ \\ \\ \\ = \boxed{2 + \sqrt{3} }$

Now ,

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

Hence,

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

(ANSWER)

Answer 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.

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

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

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

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

now,

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

rationalise the denominator.

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

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

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

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

Hence,

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

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

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