Suppose $p \in \mathit{R}$ and $\alpha , \beta$ are roots of 

$x^2-px+\frac{1}{2p^2}=0$, then the minimum value of  ${\alpha }^4+{\beta }^4$ is

$\alpha +\beta =p,\alpha \beta =1/2p^2$

$\Rightarrow {\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =p^2-1/p^2$

$\begin{array}{l}\therefore {\alpha }^4+{\beta }^4={\left({\alpha }^2+{\beta }^2\right)}^2-2{\alpha }^2{\beta }^2={\left(p^2-\frac{1}{p^2}\right)}^2-\frac{1}{2p^4}\\ =p^4+\frac{1}{p^4}-2-\frac{1}{2p^4}\\ =p^4+\frac{1}{2p^4}-2\\ \ge 2{\left(p^4\cdot \frac{1}{2p^4}\right)}^{1/2}-2=\sqrt{2}-2\end{array}$

                                                                     $[\because \mathrm{AM}\ge \mathrm{GM}]$

Thus, minimum possible value of  ${\alpha }^4+{\beta }^4\text{ is }\sqrt{2}-2$  which 

is attained when $p=1$