Minimum!

$x_1+x_2=-p, x_1x_2=\dfrac{-1}{2p^2}$ $\begin{aligned}x_1^4+x_2^4&=((x_1+x_2)^2-2x_1x_2)^2-2x_1^2x_2^2\\&=\underbrace{p^4+\dfrac{1}{2p^4}}_{\color{#D61F06}{\text{Applying AM-GM}}}+2\\&\geq \sqrt 2+2\end{aligned}$ Equality occurs when $p=\dfrac{1}{\sqrt[8]{2}}$ .

By Vieta's formulas, we have: $\begin{cases} x_1 + x_2 = - p \\ x_1 x_2 = - \dfrac{1}{2p^2} \end{cases}$

Then we have:

$\begin{aligned} x_1^2+x_2^2 & = (x_1+x_2)^2 - 2x_1x_2 \\ & = (-p)^2 - 2 \left(-\frac{1}{2p^2} \right) \\ & = p^2 + \frac{1}{p^2} \end{aligned}$

And that:

$\begin{aligned} x_1^4+x_2^4 & = (x_1^2+x_2^2)^2 - 2x_1^2x_2^2 \\ & = \left(p^2 + \frac{1}{p^2} \right)^2 - 2 \left(-\frac{1}{2p^2} \right)^2 \\ & = p^4 + 2 + \frac{1}{p^4} - \frac{1}{2p^4} \\ & = \color{#3D99F6}{p^4 + \frac{1}{2p^4}} + 2 \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{By AM-GM inequality}: p^4 + \frac{1}{2p^4} \ge 2 \sqrt{\frac{p^4}{2p^4}} = \sqrt{2}} \\ & \ge 2 + \sqrt{2} \, \approx \boxed{3.414} \end{aligned}$