Question

When denominator is rationalised, then the number $\frac{2+\sqrt{3}}{2-\sqrt{3}}$ becomes?

• A. $7-4\sqrt{3}$
• B. $3+4\sqrt{7}$
• C. $7+4\sqrt{3}$
• D. $3-4\sqrt{7}$

Answer 1

The correct option is C $7+4\sqrt{3}$

Given, $\frac{2+\sqrt{3}}{2-\sqrt{3}}$
Rationalizing the denominator,

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

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

$=7+4\sqrt{3}$