Let α, β be the roots of the equation x^2+2 √2 x-1=0. The quadratic equation, whose roots are α^4+β^4 and 1 10 (α^6+β^6 ), is :

$\begin{aligned} & x^2+2 \sqrt{2} x-1=0 \\ & \alpha+\beta=-2 \sqrt{2} \\ & \alpha \beta=-1 \\ & \alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2 \\ & =\left((\alpha+\beta)^2-2 \alpha \beta\right)^2-2(\alpha \beta)^2 \\ & =(8+2)^2-2(-1)^2 \\ & =100-2=98 \\ & \alpha^6+\beta^6=\left(\alpha^3+\beta^3\right)^2-2 \alpha^3 \beta^3 \\ & =\left((\alpha+\beta)\left((\alpha+\beta)^2-3 \alpha \beta\right)^2-2(\alpha \beta)^3\right.\end{aligned}$​x2+22​x−1=0α+β=−22​αβ=−1α4+β4=(α2+β2)2−2α2β2=((α+β)2−2αβ)2−2(αβ)2=(8+2)2−2(−1)2=100−2=98α6+β6=(α3+β3)2−2α3β3=((α+β)((α+β)2−3αβ)2−2(αβ)3​ $\begin{aligned} & =(-2 \sqrt{2}(8+3))^2+2 \\ & =(8)(121)+2=970 \\ & \frac{1}{10}\left(\alpha^6+\beta^6\right)=97 \\ & x^2-(98+97) x+(98)(97)=0 \\ & \Rightarrow x^2-195 x+9506=0\end{aligned}$​=(−22​(8+3))2+2=(8)(121)+2=970101​(α6+β6)=97x2−(98+97)x+(98)(97)=0⇒x2−195x+9506=0​