Question

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

Answer 1

It is given that;

.  ${\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}+1}=x+y\sqrt{3}}$  we need to find x and y

We know that rationalization factor for  ${\displaystyle \sqrt{3}+1}$ is${\displaystyle \sqrt{3}-1}$ . We will multiply numerator and denominator of the given expression ${\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}+1}}$by,${\displaystyle \sqrt{3}-1}$ to get

${\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{{\left(\sqrt{3}\right)}^2+{\left(1\right)}^2-2\times \sqrt{3}\times 1}{{\left(\sqrt{3}\right)}^2-{\left(1\right)}^2}}$ 

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

${\displaystyle =\frac{4-2\sqrt{3}}{2}}$

${\displaystyle =2-\sqrt{3}}$

On equating rational and irrational terms, we get 

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

Hence, we get ${\displaystyle x=2,y=-1}$