Question

Rationalize the denominator.

(i) $\frac{1}{\sqrt{5}}$ (ii) $\frac{2}{3\sqrt{7}}$ (iii) $\frac{1}{\sqrt{3}-\sqrt{2}}$ (iv) $\frac{1}{3\sqrt{5}+2\sqrt{2}}$ (v) $\frac{12}{4\sqrt{3}-\sqrt{2}}$

Answer 1

(i) $\frac{1}{\sqrt{5}}$

$$\begin{gathered} =\frac{1}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}} \\ =\frac{\sqrt{5}}{5} \end{gathered}$$

(ii) $\frac{2}{3\sqrt{7}}$

$$\begin{gathered} =\frac{2}{3\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}} \\ =\frac{2\sqrt{7}}{3\times 7} \\ =\frac{2\sqrt{7}}{21} \end{gathered}$$

(iii) $\frac{1}{\sqrt{3}-\sqrt{2}}$

$$\begin{gathered} =\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)}\times \frac{\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)} \\ =\frac{\left(\sqrt{3}+\sqrt{2}\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}\left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{\left(\sqrt{3}+\sqrt{2}\right)}{3-2} \\ =\sqrt{3}+\sqrt{2} \end{gathered}$$

(iv) $\frac{1}{3\sqrt{5}+2\sqrt{2}}$

$$\begin{gathered} =\frac{1}{\left(3\sqrt{5}+2\sqrt{2}\right)}\times \frac{\left(3\sqrt{5}-2\sqrt{2}\right)}{\left(3\sqrt{5}-2\sqrt{2}\right)} \\ =\frac{\left(3\sqrt{5}-2\sqrt{2}\right)}{{\left(3\sqrt{5}\right)}^2-{\left(2\sqrt{2}\right)}^2}\left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{\left(3\sqrt{5}-2\sqrt{2}\right)}{45-8} \\ =\frac{3\sqrt{5}-2\sqrt{2}}{37} \end{gathered}$$

(v) $\frac{12}{4\sqrt{3}-\sqrt{2}}$

$$\begin{gathered} =\frac{12}{\left(4\sqrt{3}-\sqrt{2}\right)}\times \frac{\left(4\sqrt{3}+\sqrt{2}\right)}{\left(4\sqrt{3}+\sqrt{2}\right)} \\ =\frac{12\left(4\sqrt{3}+\sqrt{2}\right)}{{\left(4\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}\left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{12\left(4\sqrt{3}+\sqrt{2}\right)}{48-2} \\ =\frac{12\left(4\sqrt{3}+\sqrt{2}\right)}{46} \\ =\frac{6\left(4\sqrt{3}+\sqrt{2}\right)}{23} \end{gathered}$$