Question

find the value of a and b
$\frac{5 + \sqrt{3} }{7 - 4 \sqrt{3 } } = 47a + \sqrt{3 \ \: b}$
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Answer 1

Answer:

Step-by-step explanation:

BRAINLIEST BRAINLIEST BRAINLIEST

Answer 2

Answer:

a = 1 and b = 27

Step-by-step explanation:

Given that -

$\implies \sf \dfrac{5 + \sqrt{3} }{7 - 4 \sqrt{3} } = 47a + \sqrt{3} b$

Consider LHS:

$\implies \sf \dfrac{5 + \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ \implies \sf \dfrac{5 + \sqrt{3} }{7 - 4 \sqrt{3} } \times \frac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} } \\ \\ \implies \sf \frac{(5 + \sqrt{3})(7 + 4 \sqrt{3})}{(7) {}^{2} - (4 \sqrt{3} ) {}^{2} } \\ \\ \implies \sf \frac{35 + 20 \sqrt{3} + 7 \sqrt{3} + 12 }{49 - 48} \\ \\ \implies \sf \frac{47 + 17 \sqrt{3} }{1} \\ \\ \implies \sf 47 + 27 \sqrt{3}$

On comparing LHS with RHS,

$\implies \sf 47 + 27 \sqrt{3} = 47a + \sqrt{3} b \\ \\ \boxed{\bf \therefore \: a = 1 \: \: and \: \: b = 27}$