Again minimum

$\alpha=a , \beta=b$

Using Vieta's,

$a + b = -p , ab = \dfrac{-1}{2p^2}$

$p^2 = a^2 + b^2 + 2ab$

$a^2 + b^2 = p^2 + \dfrac{1}{p^2} , \implies a^4 + b^4 + 2a^2b^2 = p^4 + \dfrac{1}{p^4} + 2$

$a^4 + b^4 = p^4 + \dfrac{1}{2p^4} + 2$

Even power thus it will remain as positve real

$A.M \geq G.M$

$\quad \dfrac{p^4+\dfrac {1}{2p^4}}{2} \geq \sqrt{\frac {1}{2}} = \sqrt{2}$

$\quad \Rightarrow \alpha^4 + \beta^4 = p^4 + 2 + \dfrac {1}{2p^4} \ge \boxed {2+\sqrt{2}}$

Since $\alpha$ and $\beta$ are roots of $x^2+px-\dfrac {1}{2p^2} = 0$ , using Vieta Formulas, we have:

$\quad \alpha + \beta = - p\quad \quad \alpha\beta = -\dfrac {1}{2p^2}$

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

$\quad \Rightarrow \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha \beta)$ $\quad \quad \quad \quad \quad \quad = (-p)( p^2 + \dfrac {1}{p^2}) + \dfrac {1}{2p^2} = - p^3 - \dfrac {3}{2p}$

$\quad \Rightarrow \alpha^4 + \beta^4 = (\alpha + \beta)(\alpha^3 + \beta^3) - (\alpha \beta)(\alpha^2 + \beta^2)$ $\quad \quad \quad \quad \quad \quad = p^4 + 2 + \dfrac {1}{2p^4}$

Using Cauchy-Schwart inequality, we note that:

$\quad p^4+\dfrac {1}{2p^4} \ge 2 \sqrt{\frac {1}{2}} = \sqrt{2}$

$\quad \Rightarrow \alpha^4 + \beta^4 = p^4 + 2 + \dfrac {1}{2p^4} \ge \boxed {2+\sqrt{2}}$

more accurate if 3.5 is the answer..