Question

Find the value of $a$ and $b$ in $\frac{2+\sqrt{3}}{2-\sqrt{3}}=a+b\sqrt{3}$

Answer 1

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

Rationalizing with $(2+\sqrt{3})$ be get :-

$\frac{(2+\sqrt{3})\times (2+\sqrt{3})}{(2-\sqrt{3})\times (2+\sqrt{3})}$ $=\frac{(2+\sqrt{3}{)}^2}{4-3}=4+3+4\sqrt{3}$ $=7+4\sqrt{3}$

comparing with $a+b\sqrt{3}$

we get : $a=7$ & $b=4$