If x=frac{sqrt{3}+sqrt{2}}{sqrt{3}-sqrt{2}} a | vedclass

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Question

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

Vedclass · Accepted Answer

$98$

Vedclass · Answer

$x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} 	imes \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

$=\frac{(\sqrt{3}+\sqrt{2})^{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}=\frac{3+2+2 \sqrt{3 	imes 2}}{3-2}$

$\Rightarrow \quad x=\frac{5+2 \sqrt{6}}{1}=5+2 \sqrt{6}$

Similarly, $y=5-2 \sqrt{6}$

Now, $x+y=5+2 \sqrt{6}+5-2 \sqrt{6}=10$

And, $\quad x y=\frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})} 	imes \frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})}=1$

$	herefore \quad x^{2}+y^{2}=(10)^{2}-(1)^{2}=100-2=98$

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 $x^{2}+y^{2}$.

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