Question

Q.1. Find the value of a and b
$\frac{\sqrt{3}-1}{\sqrt{3}+1}=a+b\sqrt{3}$

Answer 1

Dear student
$$\begin{gathered} \frac{\sqrt{3}-1}{\sqrt{3}+1}=\mathrm{a}+\mathrm{b}\sqrt{3} \\ \Rightarrow \frac{{\displaystyle \sqrt{3}-1}}{{\displaystyle \sqrt{3}+1}}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\mathrm{a}+\mathrm{b}\sqrt{3} \\ \Rightarrow \frac{{\displaystyle {\left(\sqrt{3}-1\right)}^2}}{{\displaystyle {\left(\sqrt{3}\right)}^2}{\displaystyle -}{\displaystyle {{\displaystyle 1}}^2}}=\mathrm{a}+\mathrm{b}\sqrt{3} \\ \Rightarrow \frac{3+1-2\sqrt{3}}{3-1}=\mathrm{a}+\mathrm{b}\sqrt{3} \\ \Rightarrow \frac{4-2\sqrt{3}}{2}=\mathrm{a}+\mathrm{b}\sqrt{3} \\ \Rightarrow 2-\sqrt{3}=\mathrm{a}+\mathrm{b}\sqrt{3} \\ \mathrm{On}\mathrm{comparing},\mathrm{we}\mathrm{get} \\ \mathrm{a}=2\mathrm{and}\mathrm{b}=-1 \end{gathered}$$
Regards