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1. If $\frac{4 + 3 \sqrt{3}}{\sqrt{7 + 4 \sqrt{3}}} = A + \sqrt{B},$ then B - A is = ?

A. -13

B. $2 \sqrt{13}$

C. 13

D. $3 \sqrt{3} - \sqrt{7}$

Solution:

\begin{aligned} \frac{4 + 3 \sqrt{3}}{\sqrt{7 + 4 \sqrt{3}}} = A + \sqrt{B} \\ \Rightarrow \sqrt{7 + 4 \sqrt{3}} \\ \Rightarrow \sqrt{2^{2} + {(\sqrt{3})}^{2} + 2 \times 2 \sqrt{3}} \\ \Rightarrow \sqrt{{(2 + \sqrt{3})}^{2}} \\ \Rightarrow (2 + \sqrt{3}) \\ \Rightarrow \frac{4 + 3 \sqrt{3}}{2 + \sqrt{3}} = A + \sqrt{B} \\ \Rightarrow \frac{4 + 3 \sqrt{3}}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = A + \sqrt{B} \\ \Rightarrow \frac{(4 + 3 \sqrt{3}) (2 - \sqrt{3})}{4 - 3} = A + \sqrt{B} \\ \Rightarrow 8 - 4 \sqrt{3} + 6 \sqrt{3} - 9 = A + \sqrt{B} \\ \Rightarrow 2 \sqrt{3} - 1 = A + \sqrt{B} \\ ∴ A = - 1 & \sqrt{B} = 2 \sqrt{3} \\ ∵ B = 2 \sqrt{3} \times 2 \sqrt{3} = 12 \\ So, B - A = 12 - (- 1) = 13 \end{aligned}

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