Question

If a and b are rational numbers and $\frac{2+\sqrt{3}}{2-\sqrt{3}}=a+b\sqrt{3}$, find the values of a and b.

Answer 1

The rationalising factor for $\frac{2+\sqrt{3}}{2-\sqrt{3}}$ is $2+\sqrt{3}$. So
$$\begin{gathered} \frac{{\left(2+\sqrt{3}\right)}^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=a+b\sqrt{3} \\ \Rightarrow \frac{4+3+4\sqrt{3}}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2}=a+b\sqrt{3} \\ \Rightarrow \frac{7+4\sqrt{3}}{4-3}=a+b\sqrt{3} \\ \Rightarrow 7+4\sqrt{3}=a+b\sqrt{3} \end{gathered}$$
Hence, a = 7 and b = 4.