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)

$(-1,12)$

(C) Given, $\frac{4+3\sqrt{3}}{\sqrt{7+4\sqrt{3}}}=a+\sqrt{b}$
$\Rightarrow a+\sqrt{b}=\frac{4+3\sqrt{3}}{\sqrt{7+2\sqrt{12}}}$
$=\frac{4+3\sqrt{3}}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$
$=-1+\sqrt{12}$

$\therefore a=-1$ and $b=12$