Question

When denominator is rationalised, then the number $\frac{4\sqrt{3}+5\sqrt{2}}{\sqrt{48}+\sqrt{18}}$ becomes?

• A. $\frac{9+4\sqrt{6}}{9}$
• B. $\frac{9-4\sqrt{6}}{15}$
• C. $\frac{9+4\sqrt{6}}{15}$
• D. $\frac{9-4\sqrt{6}}{9}$

Answer 1

Answer: (C)

$\frac{9+4\sqrt{6}}{15}$

Given, $\frac{4\sqrt{3}+5\sqrt{2}}{\sqrt{48}+\sqrt{18}}$

Rationalizing the denominator, we get

$=\frac{(4\sqrt{3}+5\sqrt{2})(4\sqrt{3}-3\sqrt{2})}{(4\sqrt{3}+3\sqrt{2})(4\sqrt{3}-3\sqrt{2})}$

$=\frac{48-12\sqrt{6}+20\sqrt{6}-30}{30}$

$=\frac{18+8\sqrt{6}}{30}$

$=\frac{9+4\sqrt{6}}{15}$