If (sqrt{3}+1)^{2}=x+sqrt{3} y, then the valu | vedclass

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(C) Given the equation: $(\sqrt{3}+1)^{2} = x + \sqrt{3}y$Expand the left side using the identity $(a+b)^2 = a^2 + 2ab + b^2$:$(\sqrt{3})^2 + 2(\sqrt{

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$6$

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(C) Given the equation: $(\sqrt{3}+1)^{2} = x + \sqrt{3}y$Expand the left side using the identity $(a+b)^2 = a^2 + 2ab + b^2$:$(\sqrt{3})^2 + 2(\sqrt{3})(1) + (1)^2 = x + \sqrt{3}y$$3 + 2\sqrt{3} + 1 = x + \sqrt{3}y$$4 + 2\sqrt{3} = x + \sqrt{3}y$By comparing the rational and irrational parts on both sides:$x = 4$$y = 2$Therefore,the value of $(x+y) = 4 + 2 = 6$.

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

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