Question

If (√3 -1 )/(√3+1 ) = x + y√3, find the value of x and y.

Answer 1

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

$- > \frac{ \sqrt{3} - 1 \times \sqrt{3} - 1}{ \sqrt{3} + 1 \times \sqrt{3} - 1}$

$- > \frac{ ({ \sqrt{3} - 1)}^{2} }{ ({ \sqrt{3)} }^{2} - {(1)}^{2} }$

$- > \frac{ ({ \sqrt{3} )}^{2} - 2( \sqrt{3} )(1) + {1}^{2} }{3 - 1}$

$- > \frac{3 - 2 \sqrt{3} + 1 }{2}$

$- > \frac{4 - 2 \sqrt{3} }{2}$

$- > \frac{2(2 - \sqrt{3)} }{2}$

$- > 2 - \sqrt{3} = x + y \sqrt{3}$

So , x = 2 , y = -1

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Answer 2

Question :-

If $\rm{ \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 } = x + y \sqrt{3} }$ , find the value of x and y.

Solution :-

Here, firstly we rationalize the denominator of LHS. Then we compare both will get the solution of x amd y.

On Rationalizing the denominator of LHS,

$\sf:\longmapsto{ \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1} \times \frac{ \sqrt{3} - 1}{ \sqrt{3} - 1} } \\ \\ \sf:\longmapsto{ \frac{( \sqrt{3} - 1) {}^{2} }{( \sqrt{3} + 1) ( \sqrt{3} - 1) } }$

In numerator using formula of (a-b)² = a² - 2ab + b² and in denominator using formula of (a+b) (a-b) = a² - b².

$\sf:\longmapsto{ \frac{( \sqrt{3} ) {}^{2} - 2( \sqrt{3} )(1) + (1) {}^{2} }{( \sqrt{3} ) {}^{2} - (1) {}^{2} } } \\ \\ \sf:\longmapsto{ \frac{3 - 2 \sqrt{3} + 1}{3 - 1} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{ \frac{ \cancel2(2 - \sqrt{3} )}{ \cancel2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto \boxed{ \sf{{2 - \sqrt{3} }}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Comparing both sides,

$\sf:\longmapsto{2 - \sqrt{3} = x + y \: \sqrt{3} } \\ \\ \bf:\longmapsto \red{x = 2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf:\longmapsto \red{y = - 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Answer :

• x = 2
• y = -1

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