Question

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

Answer 1

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

To rationalise the denominator, multiply and divide with $\sqrt{3}-1$

$\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=a+b\sqrt{3}$

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

$\frac{3+1-2\sqrt{3}}{2}=a+b\sqrt{3}$

$2-\sqrt{3}=a+b\sqrt{3}$

$\therefore a=2$ $b=-1$