Question

If $x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}},y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$, find the value of $3x^2-5xy+3y^2$

Answer 1

We have $x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=5–2\sqrt{6}$

and $y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=5+2\sqrt{6}$

Thus $x^2=(5–2\sqrt{6}{)}^2=49–4\sqrt{6}$, $y^2=(5–2\sqrt{6}{)}^2=49+4\sqrt{6}$ and $xy=1$

Therefore, $3x^2–5xy+3y^2=3\times (49–4\sqrt{6})–5+3\times (49+4\sqrt{6})=289$