Let $x^4=y$. Then we have $$y^2+ay+1=0$$$$\implies y = \frac{-a \pm \sqrt{a^2-4}}{2}$$$$\implies x = \pm \sqrt[4]{ \frac{-a \pm \sqrt{a^2-4}}{2}}$$. Note that we only have $4$ real roots of $\frac{-a - \sqrt{a^2-4}}{2}$ and $a^2-4$ are positive so $-a\ge \sqrt{a^2-4}$ and $a^2\ge 4$ implying that $a$ is negative and that $a\leq -2$. Then, we have $$-\sqrt[4]{\frac{-a+\sqrt{a^2-4}}{2}} + \sqrt[4]{\frac{-a-\sqrt{a^2-4}}{2}} = - 2\sqrt[4]{\frac{-a-\sqrt{a^2-4}}{2}}$$$$\iff \sqrt[4]{\frac{-a+\sqrt{a^2-4}}{2}} = 3\sqrt[4]{\frac{-a-\sqrt{a^2-4}}{2}}$$$$\iff \frac{-a+\sqrt{a^2-4}}{2} = \frac{81(-a-\sqrt{a^2-4})}{2}$$$$\iff -a+\sqrt{a^2-4} = -81a-81\sqrt{a^2-4}$$$$\iff 80a = -82\sqrt{a^2-4}$$$$\iff a = \frac{-41}{40}\sqrt{a^2-4}$$$$\implies a^2 = \frac{1681}{1600}\cdot (a^2-4)$$$$\implies \frac{1681}{400} = \frac{81a^2}{1600}$$$$\implies 41^2\cdot 2^2=81a^2$$$$\implies a=\pm \sqrt{\frac{82^2}{9^2}}$$$$\implies a = \pm \frac{82}{9}$$. So, we have $a=\boxed{-\frac{82}{9}}$