If $\alpha ,\beta $ are the roots of the equation ${ x }^{ 2 }+px-\cfrac { 1 }{ 2{ p }^{ 2 } } =0$, where $p\in R$. Then, the minimum value of ${ \alpha }^{ 4 }+{ \beta }^{ 4 }$ is

Roots alpha, beta satisfy alpha+beta = -p and alpha*beta = -1/(2p^2). alpha^4 + beta^4 = (alpha^2 + beta^2)^2 - 2(alpha*beta)^2 = ((alpha+beta)^2 - 2alpha*beta)^2 - 2(alpha*beta)^2. Substituting: (p^2 + 1/p^2)^2 - 2(1/(4p^4)) = p^4 + 2 + 1/p^4 - 1/(2p^4) = p^4 + 1/(2p^4) + 2. By AM-GM, p^4 + 1/(2p^4) >= 2*sqrt(1/2) = sqrt(2). Thus, min value is 2 + sqrt(2).