Question

If $a=\frac{1}{2-\sqrt{3}},b=\frac{1}{2+\sqrt{3}}$, find the value of $7a^2+11ab-7b^2$

Answer 1

We have $a=\frac{1}{2-\sqrt{3}}=\frac{1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}=2+\sqrt{3}$

and $b=\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}=2-\sqrt{3}$

Thus $a^2=(2+\sqrt{3}{)}^2=7+4\sqrt{3}$, $b^2=(2–\sqrt{3}{)}^2=7-4\sqrt{3}$ and $ab=1$

Therefore, $7a^2+11xy-7y^2=7\times (7+4\sqrt{3})+11-7\times (7-4\sqrt{3})=11+56\sqrt{3}$