Question

Can anybody solve this ??

Answer 1

$\frac{3 + 2 \sqrt{3} }{3 - 2 \sqrt{3} } \\ = \frac{3 + 2 \sqrt{3} }{3 - 2 \sqrt{3} } \times \frac{3 + 2 \sqrt{3} }{3 + 2 \sqrt{3} } \\ = \frac{(3 + 2 \sqrt{3})^{2} }{9 - 12} \\ = \frac{9 + 12 \sqrt{3} + 12}{ - 3} \\ = - 3 - 4 - 4 \sqrt{3} \\ = - 7 - 4 \sqrt{3}$

Now,

$- 7 - 4 \sqrt{3} = a + b \sqrt{3}$

By comparing,

$a = - 7 \\ b = - 4$