If $\alpha = \beta$, then $\alpha=\beta=1$, or $$1^4+k=3\times 1^2\implies k=2$$ When $\alpha \ne \beta$, then it is easy to show $\alpha^2 \ne \beta^2$. This implies $\alpha^2$ and $\beta^2$ are the two roots of $t^2 + k = 3t$ or $t^2-2t+k=0$. Therefore, by Vieta's theorem: $$ \left\{ \begin{array}{cc} \alpha^2 + \beta^2 &= 3\\ \alpha^2\cdot\beta^2 &=k \end{array} \right. \implies (\alpha+\beta)^2 = \alpha^2 + \beta^2+2\alpha\beta \implies 2^2 = 3 + 2\alpha\beta \implies \alpha\beta=\frac{1}{2}\implies k=\frac{1}{4} $$ When $k=\frac{1}{4}$, the equation becomes $x^4-3x^2+\frac{1}{4}=0$ which has two satisfactory roots $1+\frac{\sqrt{2}}{2}$ and $1-\frac{\sqrt{2}}{2}$. Therefore $k$ can be either $\boxed{2}$ or $\boxed{\frac{1}{4}}$.