Question

If $a$ and $b$ are rational numbers.
$\frac{\sqrt{3}-1}{\sqrt{3}+1}=a+b\sqrt{3}$
find $a$ and $b$.

Answer 1

Given $\frac{\sqrt{3}-1}{\sqrt{3}+1}=a+b\sqrt{3}$
First let us rationalize the denominator of $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
Multiplying numerator and denominator by $\sqrt{3}-1$, we get
$=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}$
$=\frac{3-\sqrt{3}-\sqrt{3}+1}{(\sqrt{3}{)}^2-(1{)}^2}$ [Since, $(a+b)(a-b)=a^2-b^2$]
$=\frac{3-2\sqrt{3}+1}{3-1}$
$=\frac{4-2\sqrt{3}}{2}$
$=\frac{2(2-\sqrt{3})}{2}$
$=2-\sqrt{3}$
Thus, given expression can be rewritten as
$2-\sqrt{3}=a+b\sqrt{3}$
then by principle of homoginity,
We have $a=2$ and $b=-1$