Find the values of \(x\) and \(y\)
\[ \frac{\sqrt{3} – 1}{\sqrt{3} + 1} = x + y\sqrt{3} \]
Solution:
\[ \frac{\sqrt{3} – 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} – 1}{\sqrt{3} – 1} \]
\[ = \frac{(\sqrt{3} – 1)^2}{3 – 1} \]
\[ = \frac{3 – 2\sqrt{3} + 1}{2} \]
\[ = \frac{4 – 2\sqrt{3}}{2} = 2 – \sqrt{3} \]
Comparing with \(x + y\sqrt{3}\)
\[ x = 2, \quad y = -1 \]