Question

If $\frac{\sqrt{3}-1}{\sqrt{3}+1}$ = a + b $\sqrt{3}$, find the values of a and b.

Answer 1

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

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

∴ $2-\sqrt{3}$ = a + b$\sqrt{3}$
On comparing the LHS and RHS, we get:
a = 2 and b = $-1$