Find $\frac {\alpha}{\beta} + \frac {\beta}{\alpha}$ if $\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$ The question:

Let $\alpha$ and $\beta$ be $2$ distinct real numbers which such that $\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$. Find the value of $\frac {\alpha}{\beta} + \frac {\beta}{\alpha}$.

This problem is seems to be related to Vieta's Theorem, but so far I have not used it. This is my working out:
\begin{align}
\frac {\alpha}{\beta} + \frac {\beta}{\alpha} & = \frac {\alpha^2}{\alpha \beta} + \frac {\beta^2}{\alpha \beta} \\
& = \frac {\alpha^2+\beta^2}{\alpha \beta}
\end{align}
Vieta's Theorem:

Let $p(x)=ax^2+bx+c$ be a quadratic polynomial with zeros $\alpha$,$~\beta$. Then
  $$\frac {-b}{a}=\alpha + \beta \\
\frac{c}{a} = \alpha \cdot \beta $$

Well, it is clear that the question wants us to find $\alpha^2 + \beta^2$ and $\alpha \beta$.
\begin{align}
\alpha^2+\beta^2 & = \alpha^2+2\alpha \beta + \beta^2 -2\alpha\beta \\
& = (\alpha+\beta)^2-2(\alpha\beta)
\end{align}

We have:
\begin{align}
\alpha^2 + 3\alpha + 1 & = \beta^2 + 3\beta + 1 \\
0 & =\alpha^2 - \beta^2+3\alpha-3\beta \\
& = (\alpha+\beta)(\alpha-\beta)+3(\alpha-\beta) \\
& = (\alpha-\beta)(\alpha+\beta+3) \\
\therefore~ \alpha + \beta & = -3\tag{reject $\alpha=\beta$}
\end{align}
This is where I am stuck because I am unable to find $\alpha\beta$ by algebraic manipulation. I have thought about trying to find $\alpha$ and $\beta$ through the quadratic formula, but it seems quite tedious, so it is a last resort. Is there a method to finish this question off?

$$\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$$
Implies that $\alpha$ and $\beta$ are the roots of $x^2 + 3x + 1$, thus:
$$x^2 + 3x + 1 = (x - \alpha)(x - \beta)$$
Expand $(x - \alpha)(x - \beta)$ to find  $-\alpha - \beta = 3$ and $\alpha\beta = 1$.
