Question

Rationalise the denominator of each of the following.
(i) $\frac{{\displaystyle 1}}{{\displaystyle \sqrt{7}}}$ (ii) $\frac{{\displaystyle \sqrt{5}}}{{\displaystyle 2\sqrt{3}}}$ (iii) $\frac{1}{2+\sqrt{3}}$
(iv) $\frac{1}{\sqrt{5}{\displaystyle -}{\displaystyle 2}}$ (v) $\frac{1}{5+3\sqrt{2}}$ (vi) $\frac{1}{\sqrt{7}{\displaystyle -}{\displaystyle \sqrt{6}}}$
(vii) $\frac{4}{\sqrt{11}{\displaystyle -}{\displaystyle \sqrt{7}}}$ (viii) $\frac{1+\sqrt{2}}{2-\sqrt{2}}$ (ix) $\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$

Answer 1

(i) $\frac{{\displaystyle 1}}{{\displaystyle \sqrt{7}}}$
On multiplying the numerator and denominator of the given number by $\sqrt{7}$, we get:

$\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}$

(ii) $\frac{{\displaystyle \sqrt{5}}}{{\displaystyle 2\sqrt{3}}}$
On multiplying the numerator and denominator of the given number by $\sqrt{3}$, we get:

$\frac{\sqrt{5}}{2\sqrt{3}}=\frac{\sqrt{5}}{2\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{15}}{6}$

(iii) $\frac{1}{2+\sqrt{3}}$
On multiplying the numerator and denominator of the given number by $2-\sqrt{3}$, we get:
$$\begin{gathered} \frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}} \\ =\frac{2-\sqrt{3}}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2} \\ =\frac{2-\sqrt{3}}{4-3} \\ =\frac{2-\sqrt{3}}{1} \\ =2-\sqrt{3} \end{gathered}$$

(iv) $\frac{1}{\sqrt{5}{\displaystyle -}{\displaystyle 2}}$
On multiplying the numerator and denominator of the given number by $\sqrt{5}+2$, we get:
$$\begin{gathered} \frac{1}{\sqrt{5}-2}=\frac{1}{\sqrt{5}-2}\times \frac{\sqrt{5}+2}{\sqrt{5}+2} \\ =\frac{\sqrt{5}+2}{{\left(\sqrt{5}\right)}^2-{\left(2\right)}^2} \\ =\frac{\sqrt{5}+2}{5-4} \\ =\frac{\sqrt{5}+2}{1} \\ =\sqrt{5}+2 \end{gathered}$$

(v) $\frac{1}{5+3\sqrt{2}}$
On multiplying the numerator and denominator of the given number by $5-3\sqrt{2}$, we get:
$$\begin{gathered} \frac{1}{5+3\sqrt{2}}=\frac{1}{5+3\sqrt{2}}\times \frac{5-3\sqrt{2}}{5-3\sqrt{2}} \\ =\frac{5-3\sqrt{2}}{{\left(5\right)}^2-{\left(3\sqrt{2}\right)}^2} \\ =\frac{5-3\sqrt{2}}{25-18} \\ =\frac{5-3\sqrt{2}}{7} \end{gathered}$$

(vi) $\frac{1}{\sqrt{7}{\displaystyle -}{\displaystyle \sqrt{6}}}$
Multiplying the numerator and denominator by $\sqrt{7}+\sqrt{6}$, we get
$$\begin{gathered} \frac{1}{\sqrt{7}{\displaystyle -}{\displaystyle \sqrt{6}}}=\frac{1}{\sqrt{7}{\displaystyle -}{\displaystyle \sqrt{6}}}\times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}{\displaystyle +}{\displaystyle \sqrt{6}}} \\ =\frac{\sqrt{7}+\sqrt{6}}{{\left(\sqrt{7}\right)}^2{\displaystyle -{\left(\sqrt{6}\right)}^2}} \end{gathered}$$
$$\begin{gathered} =\frac{\sqrt{7}+\sqrt{6}}{{\displaystyle 7-6}} \\ =\sqrt{7}+\sqrt{6} \end{gathered}$$

(vii) $\frac{4}{\sqrt{11}{\displaystyle -}{\displaystyle \sqrt{7}}}$
Multiplying the numerator and denominator by $\sqrt{11}+\sqrt{7}$, we get
$$\begin{gathered} \frac{4}{\sqrt{11}{\displaystyle -}{\displaystyle \sqrt{7}}}=\frac{4}{\sqrt{11}{\displaystyle -}{\displaystyle \sqrt{7}}}\times \frac{\sqrt{11}+\sqrt{7}}{\sqrt{11}{\displaystyle +}{\displaystyle \sqrt{7}}} \\ =\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{{\left(\sqrt{11}\right)}^2{\displaystyle -}{\displaystyle {\left(\sqrt{7}\right)}^2}} \end{gathered}$$
$$\begin{gathered} =\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{11{\displaystyle -}{\displaystyle 7}} \\ =\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{4} \\ =\sqrt{11}+\sqrt{7} \end{gathered}$$

(viii) $\frac{1+\sqrt{2}}{2-\sqrt{2}}$
Multiplying the numerator and denominator by $2+\sqrt{2}$, we get
$$\begin{gathered} \frac{1+\sqrt{2}}{2-\sqrt{2}}=\frac{1+\sqrt{2}}{2-\sqrt{2}}\times \frac{2+\sqrt{2}}{2+\sqrt{2}} \\ =\frac{2+\sqrt{2}+2\sqrt{2}+2}{{\left(2\right)}^2-{\left(\sqrt{2}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{4+3\sqrt{2}}{4-2} \\ =\frac{4+3\sqrt{2}}{2} \end{gathered}$$
(ix) $\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$
Multiplying the numerator and denominator by $3-2\sqrt{2}$, we get
$$\begin{gathered} \frac{3-2\sqrt{2}}{3+2\sqrt{2}}=\frac{3-2\sqrt{2}}{3+2\sqrt{2}}\times \frac{3-2\sqrt{2}}{3-2\sqrt{2}} \\ =\frac{{\left(3-2\sqrt{2}\right)}^2}{{\left(3\right)}^2-{\left(2\sqrt{2}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{9+8-12\sqrt{2}}{9-8}\left[{\left(a-b\right)}^2=a^2+b^2-2ab\right] \\ =17-12\sqrt{2} \end{gathered}$$