Question

if √3-1/√3+1=a+b√3 then find the values of a and b​

Answer 1

GIVEN :-

• (√3 - 1)/(√3 + 1) = a + b√3.

TO FIND :-

• The value of a and b.

SOLUTION :-

$\\ : \implies \displaystyle \sf \: \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1} = a + b \sqrt{3}$

$: \implies \displaystyle \sf \: \frac{ (\sqrt{3} - 1)( \sqrt{3} - 1)}{ (\sqrt{3} + 1)( \sqrt{3} - 1)} = a + b \sqrt{3}$

$: \implies \displaystyle \sf \: \frac{ (\sqrt{3} - 1) ^{2} }{ (\sqrt{3} + 1)( \sqrt{3} - 1)} = a + b \sqrt{3}$

$\bullet \ \displaystyle \sf By \ using \ identity : (a + b)(a - b) = a^{2} - b^{2}$

$: \implies \displaystyle \sf \: \dfrac{( \sqrt{3 } )^{2} + (1)^{2} - 2 \times \sqrt{3} \times 1}{( \sqrt{3}) ^{2} - (1) ^{2} } = a + b \sqrt{3}$

$: \implies \displaystyle \sf \: \frac{3 + 1 - 2 \sqrt{3} }{ \sqrt{9} - 1} = a + b\sqrt{3}$

$: \implies \displaystyle \sf \: \frac{4 - 2 \sqrt{3} }{3 - 2} = a + b\sqrt{3}$

$: \implies \displaystyle \sf \frac{4 - 2 \sqrt{3} }{2} = a + b\sqrt{3}$

$: \implies \displaystyle \sf \: \frac{2(2 - \sqrt{3} )}{2} = a + b \sqrt{3}$

$: \implies \displaystyle \sf 2 - \sqrt{3} = a + b \sqrt{3}$

$: \implies \displaystyle \sf 2 + ( - ) \sqrt{3} = a + b \sqrt{3}$

$\bullet \ \displaystyle \sf On \ comparing \ both \ sides.$

$: \implies \underline{\boxed {\displaystyle \sf a = 2\ , \ b = 3 }}$

$\therefore \underline{\displaystyle \sf Value \ of \ a \ is \ 2 \ and \ b \ is \ -1.}$

Answer 2

$Given \: a +\sqrt{3} b$

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

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

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

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

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

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

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

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

$\pink {\therefore a +\sqrt{3} b = 2 + (-1) \sqrt{3}}$

/* Compare bothsides , we get */

$\green { a= 2 \: and \: b = -1}$

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