Question

Rationalize the denominator.

(i) $\frac{1}{\sqrt{7}+\sqrt{2}}$ (ii) $\frac{3}{2\sqrt{5}-3\sqrt{2}}$ (iii) $\frac{4}{7+4\sqrt{3}}$ (iv) $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

Answer 1

(i) $\frac{1}{\sqrt{7}+\sqrt{2}}$

$$\begin{gathered} =\frac{1}{\left(\sqrt{7}+\sqrt{2}\right)}\times \frac{\left(\sqrt{7}-\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)} \\ =\frac{\sqrt{7}-\sqrt{2}}{{\left(\sqrt{7}\right)}^2-{\left(\sqrt{2}\right)}^2}\left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{\sqrt{7}-\sqrt{2}}{7-2} \\ =\frac{\sqrt{7}-\sqrt{2}}{5} \end{gathered}$$

(ii) $\frac{3}{2\sqrt{5}-3\sqrt{2}}$

$$\begin{gathered} =\frac{3}{\left(2\sqrt{5}-3\sqrt{2}\right)}\times \frac{\left(2\sqrt{5}+3\sqrt{2}\right)}{\left(2\sqrt{5}+3\sqrt{2}\right)} \\ =\frac{3\left(2\sqrt{5}+3\sqrt{2}\right)}{{\left(2\sqrt{5}\right)}^2-{\left(3\sqrt{2}\right)}^2}\left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{3\left(2\sqrt{5}+3\sqrt{2}\right)}{20-18} \\ =\frac{3}{2}\left(2\sqrt{5}+3\sqrt{2}\right) \end{gathered}$$

(iii) $\frac{4}{7+4\sqrt{3}}$

$$\begin{gathered} =\frac{4}{\left(7+4\sqrt{3}\right)}\times \frac{\left(7-4\sqrt{3}\right)}{\left(7-4\sqrt{3}\right)} \\ =\frac{4\left(7-4\sqrt{3}\right)}{{\left(7\right)}^2-{\left(4\sqrt{3}\right)}^2}\left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{4\left(7-4\sqrt{3}\right)}{49-48} \\ =28-16\sqrt{3} \end{gathered}$$

(iv) $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

$$\begin{gathered} =\frac{\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)}\times \frac{\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)} \\ =\frac{{\left(\sqrt{5}-\sqrt{3}\right)}^2}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2}\left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{{\left(\sqrt{5}\right)}^2+{\left(\sqrt{3}\right)}^2-2\left(\sqrt{5}\right)\left(\sqrt{3}\right)}{5-2} \\ =\frac{5+3-2\sqrt{15}}{3} \\ =\frac{8-2\sqrt{15}}{3} \end{gathered}$$