Math, asked by Princeroar, 1 year ago

√3-1\√3+1 = a+b√3. find the value of a and b

Answers

Answered by Anonymous
15

\mathfrak{\large{\underline{\underline{Answer:-}}}}

\boxed{ \tt a = 2, \: b = -1}

\mathfrak{\large{\underline{\underline{Explanation:-}}}}

Given :-  \sf \dfrac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 } = a +  b\sqrt{3}

To find :- Values of a and b

Solution :-

 \sf \dfrac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 } = a +  b\sqrt{3}

Consider Left Hand Side of the equation

 \sf \dfrac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 }

First Rationalise the denominator

The rationalising factor of √3 + 1 is √3 - 1. So, multiply both numerator and denominator with rationalising factor

\sf  = \dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } \times  \dfrac{ \sqrt{3} - 1 }{ \sqrt{3} - 1 }

\sf  = \dfrac{(\sqrt{3} - 1)^{2}  }{ (\sqrt{3})^{2} -  {1}^{2}  }

[Since (x + y)(x - y) = x² - y² Here x = √3 , y = 1]

 \sf  = \dfrac{ {( \sqrt{3})}^{2} - 2( \sqrt{3})(1) +  {1}^{2}}{3 - 1}

[Since (x - y)² = x² - 2xy + y² Here x = √3 , y = 1]

\sf =  \dfrac{3 - 2 \sqrt{3} + 1}{2}

\sf =  \dfrac{4 - 2 \sqrt{3} }{2}

It can be written as

 \sf  = \dfrac{2(2 -  \sqrt{3} )}{2}

 \sf  = \dfrac{ \cancel{2} (2 -  \sqrt{3} )}{ \cancel{2} }

\sf =2 -  \sqrt{3}

 \sf \dfrac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 } = 2 -  \sqrt{3}

Now consider  \bf \dfrac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 } = a +  b\sqrt{3}

\sf 2 -  \sqrt{3} = a + b \sqrt{3}

Comparing on both sides

 \sf a = 2

 \sf b \sqrt{3} = - \sqrt{3}

 \sf b \cancel{\sqrt{3}} = - \cancel{\sqrt{3}}

 \sf b = - 1

\Huge{\boxed{ \sf a = 2, \: b = -1}}

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