Solution to a system of quadratics I am learning about a Bell State, and am trying to show that they are entangled. I believe that the required proof is to show that the system
$$\alpha_0^2+\alpha_1^2=1$$
$$\beta_0^2+\beta_1^2=1$$
$$\alpha_0\beta_0=1/\sqrt{2}$$
$$\alpha_1\beta_1=1/\sqrt{2}$$
has no solutions. I have tried various ways of rewriting each variable in terms of others etc. without success. Any hints?
EDIT: I apologize, I should have made clear. In QM probabilities can be complex - what exactly this means intuitively is unclear to me, but algebraically it means that $\alpha_i,\beta_i$ can be complex.

From (3) and (4), we have:
\begin{align*}
\beta_0^2 &= \frac{1}{2\alpha_0^2} \\
\beta_1^2 &= \frac{1}{2\alpha_1^2}
\end{align*}
Plug into (2):
\begin{align*}
\frac{1}{2\alpha_0^2} + \frac{1}{2\alpha_1^2} &= 1 \\
\frac{\alpha_1^2}{\alpha_0^2\alpha_1^2} + \frac{\alpha_0^2}{\alpha_0^2\alpha_1^2} &= 2 \\
\frac{\alpha_0^2 + \alpha_1^2}{\alpha_0^2\alpha_1^2} &= 2
\end{align*}
Use (1) in the numerator to get:
$$
\alpha_0^2 \alpha_1^2 = \frac{1}{2}
$$
Therefore:
$$
\alpha_1^2 = \frac{1}{2\alpha_0^2}
$$
Plug into (1) and multiply both sides by $\alpha_0^2$:
$$
\alpha_0^4 - \alpha_0^2 + \frac{1}{2} = 0 
$$
This is a quadratic equation for $\alpha_0^2$ with no real solutions, as $\Delta = -1 \lt 0$.
