If $\alpha$ and $\beta$ are the roots of the equation x² +px+q = 0 and ${\alpha }^4$ and ${\beta }^4$ are the roots of x² –rx+s=0, the roots of x² –4qx+2q² –r = 0 are always

We have $\mathrm{\alpha }+\mathrm{\beta }=-\mathrm{p},\mathrm{\alpha \beta }=\mathrm{q},{\mathrm{\alpha }}^4+{\mathrm{\beta }}^4=\mathrm{r}$ and ${\mathrm{\alpha }}^4{\mathrm{\beta }}^4=\mathrm{s}$
Therefore, ${\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=(\mathrm{\alpha }+\mathrm{\beta }{)}^2-2\mathrm{\alpha \beta }={\mathrm{p}}^2-2\mathrm{q},$
so that
$\begin{array}{l}\mathrm{r}={\mathrm{\alpha }}^4+{\mathrm{\beta }}^4={\left({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2\right)}^2-2{\mathrm{\alpha }}^2{\mathrm{\beta }}^2={\left({\mathrm{p}}^2-2\mathrm{q}\right)}^2-2{\mathrm{q}}^2\\ \text{ i.e., }{\left({\mathrm{p}}^2\right)}^2-4\mathrm{q}\left({\mathrm{p}}^2\right)+2{\mathrm{q}}^2-\mathrm{r}=0\end{array}$
This shows that p2 is one root of ${\mathrm{x}}^2-4\mathrm{qx}+2{\mathrm{q}}^2-\mathrm{r}=0.$
If its other root is
, we have $\mathrm{\gamma }+{\mathrm{p}}^2=4\mathrm{q}$ , i.e.,$\mathrm{\gamma }=4\mathrm{q}-{\mathrm{p}}^2$
Further the discriminant of this quadratic equation is 

 $(4\mathrm{q}{)}^2-4\left(2{\mathrm{q}}^2-\mathrm{r}\right)\equiv 8{\mathrm{q}}^2+4\left[{\left({\mathrm{p}}^2-2\mathrm{q}\right)}^2-2{\mathrm{q}}^2\right]$
                     $\equiv 4{\left({\mathrm{p}}^2-2\mathrm{q}\right)}^2\ge 0$
So that both roots, p² and $-{\mathrm{p}}^2+4\mathrm{q}$ are real. Since $\alpha$ and $\beta$ are real

 ${\mathrm{p}}^2-4\mathrm{q}\ge 0$ , i.e., $-{\mathrm{p}}^2+4\mathrm{q}\le 0$ . 
Thus the roots of ${\mathrm{x}}^2-4\mathrm{qx}+2{\mathrm{q}}^2=\mathrm{r}=0$ are positive and negative