Question

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

Answer 1

$x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

$=\frac{(\sqrt{3}+\sqrt{2}{)}^2}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}=\frac{3+2+2\sqrt{3\times 2}}{3-2}(a+b{)}^2=a^2+b^2+2ab$

$\Rightarrow x=\frac{5+2\sqrt{6}}{1}=5+2\sqrt{6}$

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

$y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}(a-b{)}^2=a^2+b^2-2ab$

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

$y=5-2\sqrt{6}$

Now, $x+y=5+2\sqrt{6}+5-2\sqrt{6}=10$

And, $xy=\frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})}\times \frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})}=1$

We know that

$(x+y{)}^2=x^2+y^2+2xy$

$x^2+y^2=(x+y{)}^2-2xy$

$\therefore x^2+y^2=(10{)}^2-(1{)}^1=100-2=98$.