Question

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it is very important
9th question​

Answer 1

$\large \bf{\red{Appropriate~ question:}}$

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

$\large \bf{\red{Solution:}}$

Rationalizing the denominator on LHS :

$: \implies \bf{ \dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } } \times \bigg( \dfrac{ \sqrt{3} - 1}{ \sqrt{3} - 1} \bigg)$

$: \implies \bf{ \dfrac{( \sqrt{3} - 1) ^{2} }{( \sqrt{3} + 1)( \sqrt{3} - 1)} }$

Using the identities :

• (a - b) (a + b) = a² - b²
• (a - b)² = a² + b² - 2ab

$: \implies \bf{ \dfrac{( \sqrt{3} ) ^{2} - 2( \sqrt{3})(1) + {(1)}^{2} }{ {( \sqrt{3} )}^{2} - {(1})^{2} } }$

$: \implies \bf{ \dfrac{3 - 2 \sqrt{3} + 1}{3 - 1} }$

$: \implies \bf{ \dfrac{4 - 2 \sqrt{3} }{2} }$

Taking out 2 as common term :

$: \implies \bf \dfrac{2(2 - \sqrt{3} )}{2}$

$: \implies \bf{ \dfrac{ \cancel2(2 - \sqrt{3}) }{ \cancel2} }$

$: \implies \bf{2 - \sqrt{3} }$

When we compare this to RHS :

$\red{ \odot} \: \: \: \bf a = 2 \\ \red { \odot} \: \: \: \bf b = 1$