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

If $$\frac{{4 + 3\sqrt 3 }}{{\sqrt {7 + 4\sqrt 3 } }} = A + \sqrt B {\text{,}}$$      then B - A is = ? Correct Answer 13

$$\eqalign{ & \frac{{4 + 3\sqrt 3 }}{{\sqrt {7 + 4\sqrt 3 } }}{\text{ = }}A + \sqrt B \cr & \Rightarrow \sqrt {7 + 4\sqrt 3 } \cr & \Rightarrow \sqrt {{2^2} + {{\left( {\sqrt 3 } \right)}^2} + 2 \times 2\sqrt 3 } \cr & \Rightarrow \sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} \cr & \Rightarrow \left( {2 + \sqrt 3 } \right) \cr & \Rightarrow \frac{{4 + 3\sqrt 3 }}{{2 + \sqrt 3 }}{\text{ = }}A + \sqrt B \cr & \Rightarrow \frac{{4 + 3\sqrt 3 }}{{2 + \sqrt 3 }} \times \frac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }} = A + \sqrt B \cr & \Rightarrow \frac{{\left( {4 + 3\sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}}{{4 - 3}} = A + \sqrt B \cr & \Rightarrow 8 - 4\sqrt 3 + 6\sqrt 3 - 9 = A + \sqrt B \cr & \Rightarrow 2\sqrt 3 - 1 = A + \sqrt B \cr & \therefore A = - 1\,\,\& \,\,\sqrt B = 2\sqrt 3 \cr & \because B = 2\sqrt 3 \times 2\sqrt 3 = 12 \cr & {\text{So}},B - A = 12 - \left( { - 1} \right) = 13 \cr} $$
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