Question

Find the values of a and b in each of the following.
$\frac{\sqrt{3}+1}{\sqrt{3}-1}=a+b\sqrt{3}$

Answer 1

We have:
$\frac{\sqrt{3}+1}{\sqrt{3}-1}$

Now, rationalising the denominator of the given number by multiplying both the numerator and denominator with $\sqrt{3}+1$, we get:
$$\begin{gathered} \frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1} \\ =\frac{{\left(\sqrt{3}+1\right)}^2}{{\left(\sqrt{3}\right)}^2-{\left(1\right)}^2} \\ =\frac{3+1+2\sqrt{3}}{3-1} \\ =\frac{4+2\sqrt{3}}{2} \\ =\frac{2\left(2+\sqrt{3}\right)}{2} \\ =2+\sqrt{3} \end{gathered}$$
∴ $2+\sqrt{3}$= a + b$\sqrt{3}$
So, on comparing the LHS and the RHS, we get:
a = 2 and b = 1