Question

find the value of a and b

Answer 1

Hey buddy!!

here is yr answer....


$= > \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1 } = a + b \sqrt{3} \\ \\rationalise \: the \: denominator \\ \\ = > \frac{ \sqrt{3} + 1}{ \sqrt{3} - 1} \times \frac{ \sqrt{3} + 1}{ \sqrt{3} + 1 } = a + b \sqrt{3} \\ \\ = > \frac{( \sqrt{3} + 1)^{2} }{( \sqrt{3 } - 1)( \sqrt{3} + 1) } = a + b \sqrt{3} \\ \\ = > \frac{3 + 2 \sqrt{3} + 1 }{3 - 1} = a + b \sqrt{3} \\ \\ = > \frac{4 + 2 \sqrt{3} }{2} = a + b \sqrt{3} \\ \\ = > \frac{2(2 + \sqrt{3}) }{2} = a + b \sqrt{3} \\ \\ = > 2 + \sqrt{3} = a + b \sqrt{3} \\ \\ = > 2 + 1 \sqrt{3} = a + b \sqrt{3}$

Therefore, a = 2 , b = 1


Hope it hlpz.

Answer 2

Heya friend,
Here is the answer you were looking for:

Identities used :

${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2}$

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

On rationalizing the denominator we get,

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

On comparing we get,

$2 + \sqrt{3} = a + b \sqrt{3} \\ \\ a = 2 \: : \: b = 1$

Hope this helps!!!

Feel free to ask in the comment section if you have any doubt regarding to my answer...

@Mahak24

Thanks...
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