If frac{4 + 3sqrt{3}}{sqrt{7 + 4sqrt{3}}} = a | vedclass

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(C) Given expression is $\frac{4 + 3\sqrt{3}}{\sqrt{7 + 4\sqrt{3}}}$.First,simplify the denominator $\sqrt{7 + 4\sqrt{3}}$.We know that $7 + 4\sqrt{3}

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$(-1, 12)$

Vedclass · Answer

(C) Given expression is $\frac{4 + 3\sqrt{3}}{\sqrt{7 + 4\sqrt{3}}}$.First,simplify the denominator $\sqrt{7 + 4\sqrt{3}}$.We know that $7 + 4\sqrt{3} = 7 + 2(2\sqrt{3}) = 2^2 + (\sqrt{3})^2 + 2(2)(\sqrt{3}) = (2 + \sqrt{3})^2$.So,$\sqrt{7 + 4\sqrt{3}} = \sqrt{(2 + \sqrt{3})^2} = 2 + \sqrt{3}$.Now,the expression becomes $\frac{4 + 3\sqrt{3}}{2 + \sqrt{3}}$.Rationalizing the denominator: $\frac{(4 + 3\sqrt{3})(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{8 - 4\sqrt{3} + 6\sqrt{3} - 3(3)}{4 - 3} = \frac{8 + 2\sqrt{3} - 9}{1} = -1 + 2\sqrt{3}$.We can write this as $-1 + \sqrt{2^2 	imes 3} = -1 + \sqrt{12}$.Comparing with $a + \sqrt{b}$,we get $a = -1$ and $b = 12$.Thus,$(a, b) = (-1, 12)$.

If $\frac{4 + 3\sqrt{3}}{\sqrt{7 + 4\sqrt{3}}} = a + \sqrt{b}$,then $(a, b) = $

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