Estamos considerando el campo $F_{11}(\alpha,\beta)$ %#% de fueron #% y $\alpha^2 + 5\alpha +1=0$. Vamos a ver que %#% $ #%
De hecho sumando las ecuaciones obtenemos $\beta^2 - 5\beta +2=0$$$F{11}(\alpha,\beta)=F{11}(\alpha+\beta)$ (\alpha + \beta)$$(\alpha + \beta)(\alpha - \beta + 5)=1$ (\alpha - \beta + 5)$ which means that these elements are inverse one another, thus adding $\alpha$ to the base field we add its inverse $\beta$ as well, and also their sum and their difference and thus both $ (\alpha + \beta) = \frac{\sqrt{21}+\sqrt{17}}{2}$ and $x^4-19 x ^ 2 +1 $. We proved the claim. Now notice that $F_ {11} $ simply solving the equations of degree two. It has minimum polynomial $x ^-19x^2+1=(x^2+5) 4 (x ^ 2-2) $. It can be factorized in $\alpha + \beta$ as $ p (x) $ thus $% $ $ is a solution of one of them, let's call it $