Question

Rationalise the denominator of $\left(\frac{5\sqrt{3}-4\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}\right)$.

Answer 1

$$\begin{gathered} \frac{5\sqrt{3}-4\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}=\frac{5\sqrt{3}-4\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}\times \frac{4\sqrt{3}-3\sqrt{2}}{4\sqrt{3}-3\sqrt{2}} \\ =\frac{\left(5\sqrt{3}-4\sqrt{2}\right)\times \left(4\sqrt{3}-3\sqrt{2}\right)}{{\left(4\sqrt{3}\right)}^2-{\left(3\sqrt{2}\right)}^2} \\ =\frac{\left(5\sqrt{3}\times 4\sqrt{3}\right)-\left(5\sqrt{3}\times 3\sqrt{2}\right)-\left(4\sqrt{2}\times 4\sqrt{3}\right)+\left(4\sqrt{2}\times 3\sqrt{2}\right)}{{\left(4\sqrt{3}\right)}^2-{\left(3\sqrt{2}\right)}^2} \\ =\frac{20\times 3-15\sqrt{6}-16\sqrt{6}+12\times 2}{16\times 3-9\times 2} \\ =\frac{60-31\sqrt{6}+24}{48-18} \\ =\frac{84-31\sqrt{6}}{30} \end{gathered}$$