If x = sqrt{7} + sqrt{3} and xy = 4,then x^4 | vedclass

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(B) Given $x = \sqrt{7} + \sqrt{3}$ and $xy = 4$. Since $xy = 4$,we have $y = \frac{4}{x} = \frac{4}{\sqrt{7} + \sqrt{3}}$. Rationalizing the denomina

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$368$

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(B) Given $x = \sqrt{7} + \sqrt{3}$ and $xy = 4$. Since $xy = 4$,we have $y = \frac{4}{x} = \frac{4}{\sqrt{7} + \sqrt{3}}$. Rationalizing the denominator: $y = \frac{4(\sqrt{7} - \sqrt{3})}{7 - 3} = \frac{4(\sqrt{7} - \sqrt{3})}{4} = \sqrt{7} - \sqrt{3}$. Now,$x + y = (\sqrt{7} + \sqrt{3}) + (\sqrt{7} - \sqrt{3}) = 2\sqrt{7}$. We know that $x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2$. First,calculate $x^2 + y^2 = (x + y)^2 - 2xy = (2\sqrt{7})^2 - 2(4) = 28 - 8 = 20$. Then,$x^4 + y^4 = (20)^2 - 2(xy)^2 = 400 - 2(4)^2 = 400 - 2(16) = 400 - 32 = 368$.

If $x = \sqrt{7} + \sqrt{3}$ and $xy = 4$,then $x^4 + y^4 =$

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