Question

If $\frac{2+\sqrt{3}}{2-\sqrt{3}}=A+B\sqrt{3}$, then find the values of $A$ and $B$.

Answer 1

Given:

$\frac{2+\sqrt{3}}{2-\sqrt{3}}=A+B\sqrt{3}$

Lets take $\frac{2+\sqrt{3}}{2-\sqrt{3}}$ and rationalize the denominator.

Since, Rationalizing factor of $2-\sqrt{3}$ is $2+\sqrt{3}$.

Multifly both numerator and the denominator with $2+\sqrt{3}$

$\frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}$

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

$=\frac{2^2+(\sqrt{3}{)}^2+4\sqrt{3}}{(2{)}^2-(\sqrt{3}{)}^2}$

$=\frac{7+4\sqrt{3}}{4-3}$

$=7+4\sqrt{3}=A+B\sqrt{3}$

$\therefore \frac{2+\sqrt{3}}{2-\sqrt{3}}=7+4\sqrt{3}$

$\Rightarrow A+B\sqrt{3}=7+4\sqrt{3}$

By comparing the corresponding parts of rational and irrationals numbers,

$A=7,\text{and}B=4$.