Question

If $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ find $x^2+\frac{1}{x^2}$

Answer 1

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

$=\frac{(\sqrt{3}+\sqrt{2}{)}^2}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}=\frac{3+2\sqrt{6}+2}{3-2}$

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

$\frac{1}{x}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{(\sqrt{3}-\sqrt{2}{)}^2}{{\sqrt{3}}^2-{\sqrt{2}}^2}=5-2\sqrt{6}$

$\therefore x+\frac{1}{x}=10$

Squaring both sidees:

$x^2+2x-\frac{1}{x}+\frac{1}{x^2}=100\Rightarrow x^2+\frac{1}{x^2}+2=100$

$\Rightarrow x^2+\frac{1}{x^2}=98$