If 4 + 3 3 7 + 4 3 = A + B , then B - A is

Answer: Option C

$$\begin{array}{rl}& \frac{4+3\sqrt{3}}{\sqrt{7+4\sqrt{3}}}=\text{A}+\sqrt{\text{B}}\\ & \Rightarrow \sqrt{7+4\sqrt{3}}\\ & \Rightarrow \sqrt{{\left(2\right)}^2+{\left(\sqrt{3}\right)}^2+2\times 2\sqrt{3}}\\ & \Rightarrow \sqrt{{\left(2+\sqrt{3}\right)}^2}\\ & \Rightarrow \left(2+\sqrt{3}\right)\\ & \Rightarrow \frac{4+3\sqrt{3}}{2+\sqrt{3}}=\text{A}+\sqrt{\text{B}}\\ & \Rightarrow \frac{4+3\sqrt{3}}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}=\text{A}+\sqrt{\text{B}}\\ & \Rightarrow \frac{\left(4+3\sqrt{3}\right)\left(2-\sqrt{3}\right)}{4-3}=\text{A}+\sqrt{\text{B}}\\ & \Rightarrow 8-4\sqrt{3}+6\sqrt{3}-9=\text{A}+\sqrt{\text{B}}\\ & \Rightarrow 2\sqrt{3}-1=\text{A}+\sqrt{\text{B}}\\ & \text{A}=-1\text{ and }\sqrt{\text{B}}=2\sqrt{3}\\ & \text{B}=2\sqrt{3}\times 2\sqrt{3}=12\\ & \text{B }-\text{ A}=12-\left(-1\right)=13\end{array}$$