Rationalise the denominator of each of the following:
$\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

Open in App

Solution

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

Suggest Corrections

Similar questions

\frac{1}{\sqrt{7} + \sqrt{6} - \sqrt{13}}

\frac{3}{\sqrt{3} + \sqrt{5} - \sqrt{2}}

\frac{4}{2 + \sqrt{3} + \sqrt{7}}

\frac{1}{(2 + \sqrt{3})}

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

Rationalise the denominator $\frac{1}{\sqrt{7} + \sqrt{6} - \sqrt{13}}$

\frac{1}{3 \sqrt{2} - 2 \sqrt{3}}