Question

find the value of a and b in the following 1) 5+2√3/7+4√3=a-b√3 (2) √3-1/√3+1=a+b√3

Answer 1

Heya there !!!
Here is the answer you were looking for :

Identities used :

$(x + y)(x - y) = {x}^{2} - {y}^{2} \\ {(x - y) }^{2} = {x}^{2} + {y}^{2} - 2xy$


$(1)$
$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } = a - b \sqrt{3}$

L.H.S

On rationalizing the denominator we get,

$= \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ = \frac{5(7 - 4 \sqrt{3} ) + 2 \sqrt{3}(7 - 4 \sqrt{3}) }{ {(7)}^{2} - {(4 \sqrt{3} )}^{2} } \\ \\ = \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24}{49 - 48} \\ \\ = 11 - 6 \sqrt{3}$

On comparing we get,

$11 - 6 \sqrt{3} = a - b \sqrt{3} \\ \\ a = 11 \: : \: b = 6$

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

L.H.S

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!!!

If you have any doubt regarding to my answer, please ask in the comment section ^_^

@Mahak24

Thanks...
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Answer 2

Hello dear user ...

Solution Given below ☺✌
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Question Given =
$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } = a - b \sqrt{3} \\ \\ = > \frac{(5 + 2 \sqrt{3}) }{(7 + 4 \sqrt{3} )} \times \frac{(7 - 4 \sqrt{3}) }{(7 - 4 \sqrt{3} )} = a - \sqrt{3} \\ \\ multiply \: \: nr \: \: and \: \: dr \: \: \\ \\ = > \frac{5(7 - 4 \sqrt{3}) + 2 \sqrt{3} (7 - 4 \sqrt{3} )}{( {7})^{2} - (4 \sqrt{3} {)}^{2} } \\ \\ = > \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24 }{49 - 48} \\ \\ = > \frac{11 - 6 \sqrt{3} }{1} = 11 - 6 \sqrt{3}$
iF we comparing with a - b√3

we get

a = 11 and b = 6

$a - b \sqrt{3} = 11 - 6 \sqrt{3}$
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2.

Question =
$\frac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } = a + b \sqrt{3} \\ \\ = > \frac{( \sqrt{3} - 1)}{( \sqrt{3} + 1)} \times \frac{( \sqrt{3} - 1) }{( \sqrt{3} - 1)} = a + b \sqrt{3} \\ \\ = > \frac{( \sqrt{3} - 1 {)}^{2} }{( \sqrt{3} {)}^{2} - (1 {)}^{2} } = a - b \sqrt{3} \\ \\ = > \frac{3 + 1 - 2 \sqrt{3} }{3 - 1} \\ \\ = > \frac{4 - 2 \sqrt{3} }{2} = 2(2 - \sqrt{3} ) \\ \\ = > 2 - \sqrt{3}$
If we comparing with a + b√3

we get =

a = 2 and. b = -1
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Hope it's helps you.
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