Question

Rationalize $\frac{\sqrt{3}-1}{\sqrt{3}+1}$

Answer 1

Given expression

$\frac{\sqrt{3}-1}{\sqrt{3}+1}$

We know that

${(a-b)}^2=a^2-2ab+b^2$ and $(a+b)(a-b)=a^2-b^2$

To rationalize the given expression we will multiply and divide the denominator by $\sqrt{3}–1$

$$\begin{gathered} \Rightarrow \frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{{(\sqrt{3}-1)}^2}{(\sqrt{3}+1)(\sqrt{3}-1)} \\ =\frac{3+1-2\sqrt{3}}{3-1}=\frac{4-2\sqrt{3}}{2} \\ =\frac{2(2-\sqrt{3})}{2}=2-\sqrt{3} \end{gathered}$$

Hence we get $2-\sqrt{3}$ after rationalization.