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

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Question

(C) Rationalizing the denominator for $x$:$x = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} 	imes \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} = \frac{

Vedclass · Accepted Answer

$98$

Vedclass · Answer

(C) Rationalizing the denominator for $x$:$x = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} 	imes \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} = \frac{(\sqrt{3}+\sqrt{2})^{2}}{3-2} = 3+2+2\sqrt{6} = 5+2\sqrt{6}$.Similarly,for $y$:$y = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} 	imes \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{(\sqrt{3}-\sqrt{2})^{2}}{3-2} = 3+2-2\sqrt{6} = 5-2\sqrt{6}$.Now,calculate $x+y$ and $xy$:$x+y = (5+2\sqrt{6}) + (5-2\sqrt{6}) = 10$.$xy = (5+2\sqrt{6})(5-2\sqrt{6}) = 5^{2} - (2\sqrt{6})^{2} = 25 - 24 = 1$.Using the identity $x^{2}+y^{2} = (x+y)^{2} - 2xy$:$x^{2}+y^{2} = (10)^{2} - 2(1) = 100 - 2 = 98$.

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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