Question

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

• A. $(12,-1)$
• B. $(1,12)$
• C. $(-1,12)$
• D. $(-12,1)$

Answer 1

Answer: (C)

$(-1,12)$

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

As $\sqrt{7+4\sqrt{3}}=\sqrt{2^2+2\times \left(2\right)\left(\sqrt{3}\right)+(\sqrt{3}{)}^2}$

$\Rightarrow \sqrt{7+4\sqrt{3}}=\sqrt{(2+\sqrt{3}{)}^2}=(2+\sqrt{3})$

$\therefore \frac{4+3\sqrt{3}}{2+\sqrt{3}}=\frac{(4+3\sqrt{3})(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}$

$\Rightarrow \frac{8+6\sqrt{3}-4\sqrt{3}-9}{1}=-1+2\sqrt{3}$

$\Rightarrow a+\sqrt{b}=-1+\sqrt{12}$

$\therefore (a,b)is(-1,12)$