Question

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

Answer 1

$x=\frac{\sqrt{3}+2}{\sqrt{3}-2}\times \frac{\sqrt{3}+2}{\sqrt{3}+2}=\frac{3+2+4\sqrt{3}}{3-4}=\frac{5+4\sqrt{3}}{-1}=-5-4\sqrt{3}$

$y=\frac{\sqrt{3}-2}{\sqrt{3}+2}\times \frac{\sqrt{3}-2}{\sqrt{3}-2}=\frac{3+2-4\sqrt{3}}{3-4}=\frac{5-4\sqrt{3}}{-1}=4\sqrt{3}-5$

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

$\Rightarrow x^2+y^2={(-5-4\sqrt{3}+4\sqrt{3}-5)}^2-2(-5-4\sqrt{3})(4\sqrt{3}-5)$

$\Rightarrow x^2+y^2={(-10)}^2+2(4\sqrt{3}+5)(4\sqrt{3}-5)$

$\Rightarrow x^2+y^2={(-10)}^2+2(48-25)$

$\Rightarrow x^2+y^2={(-10)}^2+2\left(23\right)$

$\Rightarrow x^2+y^2=100+46=146$

$\therefore x^2+y^2=146$