Question

If $x$ and $y$ are rational numbers such that $\frac{2+\sqrt{3}}{2-\sqrt{3}}=x+y\sqrt{3}$, find the value of $x$ and $y$.

Answer 1

We have,

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

On rationalizing the denominator, we get:

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

$\Rightarrow \frac{(2+\sqrt{3}{)}^2}{2^2-(\sqrt{3}{)}^2}=x+y\sqrt{3}$

$\Rightarrow \frac{4+3+4\sqrt{3}}{4-3}=x+y\sqrt{3}$

$\Rightarrow 7+4\sqrt{3}=x+y\sqrt{3}$

On comparing both sides, we get:

$x=7,y=4$

Hence, the values of $x$ and $y$ are $7$ and $4$, respectively.