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(C) Given equation: $x^2+2 \sqrt{2} x-1=0$Sum of roots: $\alpha+\beta = -2 \sqrt{2}$Product of roots: $\alpha \beta = -1$Calculate $\alpha^2+\beta^2 =

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$x^2-195 x+9506=0$

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(C) Given equation: $x^2+2 \sqrt{2} x-1=0$Sum of roots: $\alpha+\beta = -2 \sqrt{2}$Product of roots: $\alpha \beta = -1$Calculate $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha \beta = (-2 \sqrt{2})^2 - 2(-1) = 8+2 = 10$Calculate $\alpha^4+\beta^4 = (\alpha^2+\beta^2)^2 - 2(\alpha \beta)^2 = (10)^2 - 2(-1)^2 = 100-2 = 98$Calculate $\alpha^6+\beta^6 = (\alpha^2+\beta^2)(\alpha^4 - \alpha^2 \beta^2 + \beta^4) = (10)(98 - (-1)^2) = 10(97) = 970$Roots of the new equation are $98$ and $\frac{1}{10}(970) = 97$Sum of new roots: $98+97 = 195$Product of new roots: $98 	imes 97 = 9506$The required quadratic equation is $x^2 - (	ext{sum})x + (	ext{product}) = 0$,which is $x^2 - 195x + 9506 = 0$.

Let $\alpha, \beta$ be the roots of the equation $x^2+2 \sqrt{2} x-1=0$. The quadratic equation,whose roots are $\alpha^4+\beta^4$ and $\frac{1}{10}(\alpha^6+\beta^6)$,is :

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