Question

If $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}};y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ then find $x^2+y^2$

Answer 1

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

Rationalising the denominator, we get

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

$\Rightarrow x=\frac{3+2+2\sqrt{6}}{3-2}$

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

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

Rationalising the denominator, we get

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

$\Rightarrow y=\frac{3+2-2\sqrt{6}}{3-2}$

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

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

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

Substituting the values of $x$ and $y$ we have

$\Rightarrow x^2+y^2={(5+2\sqrt{6}+5-2\sqrt{6})}^2-2(5+2\sqrt{6})(5-2\sqrt{6})$

$={10}^2-2(25-4\times 6)$

$=100-2(25-24)$

$=100-2\left(1\right)$

$=100-2=98$

$\therefore x^2+y^2=98$