Question

If $\frac{\sqrt{3}-1}{\sqrt{3}+1}=x+y\sqrt{3}$, find the value of x and y.

Answer 1

$\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}$

(Rationalising the denominator)

$=\frac{(\sqrt{3}-1{)}^2}{(\sqrt{3}{)}^2-(1{)}^2}=\frac{3+1-2\sqrt{3}}{3-1}=\frac{4-2\sqrt{3}}{2}$

$=2-\sqrt{3}$

$\therefore 2-\sqrt{3}=x+y\sqrt{3}$

Comparing, we get

x = 2, y = -1