Question

$\frac{7 + 4 \sqrt {3} } {7 - 4 \sqrt{3} } = a + b \sqrt{3} find \: a \: and \: b$

Answer 1

Hey buddy, here is your answer.

We have to rationalise it,
therefore, the equation would be,

$\frac{7 + 4 \sqrt{3} }{7 - 4 \sqrt{3} } \times \frac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} } = a + b \sqrt{3}$
$\frac{49 + 28 \sqrt{3} + 28 \sqrt{3} + 48 }{(7 - 4 \sqrt{3})(7 + 4 \sqrt{3}) } = a + b \sqrt{3}$
Since, (a-b)(a+b) = (a^2-b^2)

$\frac{97 + 56 \sqrt{3} }{ {7}^{2} - {(4 \sqrt{3} )}^{2} } = a + b \sqrt{3}$
$\frac{97 + 56 \sqrt{3} }{49 - 48} = a + b \sqrt{3}$
$\frac{97 + 56 \sqrt{3} }{1} = a + b \sqrt{3}$
$\frac{97}{1} + \frac{56 \sqrt{3} }{1} = a + b \sqrt{3}$
On comparing,
we get,

a = 97
&
$b \sqrt{3} = 56 \sqrt{3} \\ therefore \: \: b = 56$
Hope it helps you.
Plz mark me as brainliest.