If x=frac{sqrt{3}+1}{sqrt{3}-1} and y=frac{sq | vedclass

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(C) First,rationalize $x$: $x = \frac{\sqrt{3}+1}{\sqrt{3}-1} 	imes \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{(\sqrt{3}+1)^{2}}{3-1} = \frac{3+1+2\sqrt{3

Vedclass · Accepted Answer

$14$

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(C) First,rationalize $x$: $x = \frac{\sqrt{3}+1}{\sqrt{3}-1} 	imes \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{(\sqrt{3}+1)^{2}}{3-1} = \frac{3+1+2\sqrt{3}}{2} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$. Next,rationalize $y$: $y = \frac{\sqrt{3}-1}{\sqrt{3}+1} 	imes \frac{\sqrt{3}-1}{\sqrt{3}-1} = \frac{(\sqrt{3}-1)^{2}}{3-1} = \frac{3+1-2\sqrt{3}}{2} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$. Now,calculate $x^{2}+y^{2}$: $x^{2}+y^{2} = (2+\sqrt{3})^{2} + (2-\sqrt{3})^{2}$. Using the identity $(a+b)^{2} + (a-b)^{2} = 2(a^{2}+b^{2})$: $x^{2}+y^{2} = 2(2^{2} + (\sqrt{3})^{2}) = 2(4+3) = 2(7) = 14$.

If $x=\frac{\sqrt{3}+1}{\sqrt{3}-1}$ and $y=\frac{\sqrt{3}-1}{\sqrt{3}+1},$ then the value of $(x^{2}+y^{2})$ is

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