Question

If a and b are rational numbers and $\frac{2+\sqrt{3}}{2-\sqrt{3}}=a+b\sqrt{3}$ find the values of a and b.

• A. a = 7, b = 4
• B. a = 4, b = 7
• C. a = 11, b = 14
• D. a = 7, b = 11

Answer 1

The correct option is A a = 7, b = 4

$\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{(2+\sqrt{3})}{(2-\sqrt{3})}\times \frac{(2+\sqrt{3})}{(2+\sqrt{3})}=\frac{(2+\sqrt{3}{)}^2}{(4-3)}=(2+\sqrt{3}{)}^2=[2^2+(\sqrt{3}{)}^2+2\times 2\times \sqrt{3}]=(4+3+4\sqrt{3})=(7+4\sqrt{3})\therefore \frac{2+\sqrt{3}}{2-\sqrt{3}}=a+\sqrt{3}\iff (7+4\sqrt{3})=(a+b\sqrt{3})$
$\iff a=7$ and b = 4