Question

If $x=\frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})}andy=\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3+\sqrt{2}}\right)},find(x^2+y^2)$

• A. 98
• B. 80
• C. 89
• D. 90

Answer 1

The correct option is A

98

$x=\frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})}\times \frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})}=\frac{(\sqrt{3}+\sqrt{2}{)}^2}{(3-2)}=(5+2\sqrt{6})y=\frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})}\times \frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3}-\sqrt{2})}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{(\sqrt{3}-\sqrt{2}{)}^2}{(3-2)}=(5-2\sqrt{6})\therefore x+y=10andxy=(25-24)=1\Rightarrow (x+y{)}^2={10}^2\Rightarrow x^2+y^2+2xy=100\Rightarrow x^2+y^2+2\times 1=100\Rightarrow x^2+y^2=98$