Question

If 5+ root 3 divided by 7-4 root 3 = a+ root 3 b then find a&b

Answer 1

ANSWER

a=71, b=27

SOLUTION

5+√3 × 7+4√3 (rationalising)

7-4√3 × 7+4√3

5+√3(7+4√3)

(7)²-(4√3)² [(a-b)(a+b)=a²-b²]

5(7+4√3)+√3(7+4√3)

49-16×3 [(√a)²=a]

35+20√3+7√3+4×9

1

20√3+7√3+71 = a+b√3 (given)

71+27√3 = a+b√3

a= 71

b√3= 27√3, b=27 (√3 and √3 cut from each other)

HENCE

a=71

b=27

Hope you got it

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Thank you!

Answer 2

Step-by-step explanation:

$\bf \underline{Given-}$

$\sf{ \frac{5 + \sqrt{3} }{7 - 4 \sqrt{3} } = a + b \sqrt{3} }$

$\bf \underline{To\: find-}$

$\textsf{the value of a and b = ?}$

$\bf \underline{Solution-}$

$\textsf{We have}$

$\sf{ \frac{5 + \sqrt{3} }{7 - 4 \sqrt{3} } }$

$\textsf{The denominator is 7 - 4√3.}$

$\textsf{We know that,}$

$\textsf{Rationalising factor of a - b√c = a + b√c.}$

$\textsf{So, the rationalising factor of 7 - 4√3 = 7 + 4√3.}$

$\textsf{On rationalising the denominator them}$

$\sf{ \implies \: \frac{5 + \sqrt{3} }{7 - 4 \sqrt{3} } \times \frac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} } }$

$\sf{ \implies \: \frac{(5 + \sqrt{3})(7 + 4 \sqrt{3} ) }{(7 - 4 \sqrt{3} )(7 + 4 \sqrt{3}) } }$

$\textsf{Now, comparing the denominator with (a-b)(a+b), we get}$

$\sf{ \: \: \: a = 7 \: and \: b = 4 \sqrt{3} }$

$\textsf{Using identity (a-b)(a+b) = a² - b², we get}$

$\sf{ \implies \: \frac{(5 + \sqrt{3} )(7 + 4 \sqrt{3} )}{(7 {)}^{2} - (4 \sqrt{3} {)}^{2} } }$

$\sf{ \implies \: \frac{(5 + \sqrt{3} )(7 + 4 \sqrt{3} )}{49 - 48} }$

$\sf{ \implies \: \frac{(5 + \sqrt{3} )(7 + 4 \sqrt{3} )}{1} }$

$\sf{ \implies \:(5 + \sqrt{3} )(7 + 4 \sqrt{3} )}$

$\sf{ \implies \:35 + 20 \sqrt{3} + 7 \sqrt{3} + 4 (\sqrt{3} {)}^{2} }$

$\sf{ \implies \:35 + 27 \sqrt{3} + 4(3)}$

$\sf{ \implies \:35 + 27 \sqrt{3} + 12 }$

$\sf{ \implies \:47 + 27 \sqrt{3} }$

$\sf{ \therefore \: 47 + 27 \sqrt{3} = a + b \sqrt{3} }$

$\textsf{On comparing with the value of RHS, we notice that}$

$\sf{a = 47 \: and \: b = 27 \sqrt{3} }$

$\bf \underline{Hence, the\: value\: of \: a\:= 47\: and\:b\: = \: 27. }$