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 $\frac{1}{10}\left(\alpha^6+\beta^6\right)$, is :

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Aicharya · Accepted Answer

$\begin{aligned} & x^2+2 \sqrt{2} x-1=0 \ & \alpha+\beta=-2 \sqrt{2} \ & \alpha \beta=-1 \ & \alpha^4+\beta^4=\left(\alpha^2+\beta^2ight)^2-2 \alpha^2 \beta^2 \ & =\left((\alpha+\beta)^2-2 \alpha \betaight)^2-2(\alpha \beta)^2 \ & =(8+2)^2-2(-1)^2 \ & =100-2=98 \ & \alpha^6+\beta^6=\left(\alpha^3+\beta^3ight)^2-2 \alpha^3 \beta^3 \ & =\left((\alpha+\beta)\left((\alpha+\beta)^2-3 \alpha \betaight)^2-2(\alpha \beta)^3ight.\end{aligned}$ $\begin{aligned} & =(-2 \sqrt{2}(8+3))^2+2 \ & =(8)(121)+2=970 \ & \frac{1}{10}\left(\alpha^6+\beta^6ight)=97 \ & x^2-(98+97) x+(98)(97)=0 \ & \Rightarrow x^2-195 x+9506=0\end{aligned}$

Let $\alpha, \beta$ be the roots of the equation $x^2+2 \sqrt{2} x-1=0$. The quadratic equation, whose roots…

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