Let x = frac{sqrt{13} + sqrt{11}}{sqrt{13} - | vedclass

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(A) Given $x = \frac{\sqrt{13} + \sqrt{11}}{\sqrt{13} - \sqrt{11}}$ and $y = \frac{1}{x} = \frac{\sqrt{13} - \sqrt{11}}{\sqrt{13} + \sqrt{11}}$.First,

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$1717$

Vedclass · Answer

(A) Given $x = \frac{\sqrt{13} + \sqrt{11}}{\sqrt{13} - \sqrt{11}}$ and $y = \frac{1}{x} = \frac{\sqrt{13} - \sqrt{11}}{\sqrt{13} + \sqrt{11}}$.First,calculate $x + y$:$x + y = \frac{\sqrt{13} + \sqrt{11}}{\sqrt{13} - \sqrt{11}} + \frac{\sqrt{13} - \sqrt{11}}{\sqrt{13} + \sqrt{11}}$$= \frac{(\sqrt{13} + \sqrt{11})^2 + (\sqrt{13} - \sqrt{11})^2}{(\sqrt{13} - \sqrt{11})(\sqrt{13} + \sqrt{11})}$$= \frac{(13 + 11 + 2\sqrt{143}) + (13 + 11 - 2\sqrt{143})}{13 - 11}$$= \frac{24 + 24}{2} = \frac{48}{2} = 24$.Next,calculate $xy$:$xy = x \cdot \frac{1}{x} = 1$.Now,evaluate $3x^2 - 5xy + 3y^2$:$3x^2 - 5xy + 3y^2 = 3(x^2 + y^2) - 5xy$$= 3((x + y)^2 - 2xy) - 5xy$$= 3(x + y)^2 - 6xy - 5xy$$= 3(x + y)^2 - 11xy$$= 3(24)^2 - 11(1)$$= 3(576) - 11$$= 1728 - 11 = 1717$.

Let $x = \frac{\sqrt{13} + \sqrt{11}}{\sqrt{13} - \sqrt{11}}$ and $y = \frac{1}{x}$. Then the value of $3x^2 - 5xy + 3y^2$ is:

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