Question

Simplify by rationalising the denominator:
$\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}}$

• A. $\frac{114-41\sqrt{6}}{30}$
• B. $\frac{114+41\sqrt{6}}{30}$
• C. $-\frac{114-41\sqrt{6}}{30}$
• D. $-\frac{114+41\sqrt{6}}{30}$

Answer 1

The correct option is A $\frac{114-41\sqrt{6}}{30}$

As given $\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}}$

$\because \sqrt{48}+\sqrt{18}=\sqrt{16\times 3}+\sqrt{9\times 2}=4\sqrt{3}+3\sqrt{2}$

$\therefore \frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}}=\frac{7\sqrt{3}-5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}$

$=\frac{7\sqrt{3}-5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}\times \frac{4\sqrt{3}-3\sqrt{2}}{4\sqrt{3}-3\sqrt{2}}$ [By rationalising]
$=\frac{7\sqrt{3}\times 4\sqrt{3}-7\sqrt{3}\times 3\sqrt{2}-5\sqrt{2}\times 4\sqrt{3}+5\sqrt{2}\times 3\sqrt{2}}{{\left(4\sqrt{3}\right)}^2-{\left(3\sqrt{2}\right)}^2}$

$=\frac{28\times 3-21\sqrt{6}-20\sqrt{6}+15\times 2}{16\times 3-9\times 2}$

$=\frac{84+30-41\sqrt{6}}{30}$

$=\frac{114-41\sqrt{6}}{30}$