Maximum likelihood estimator for beta distribution I'm trying to derive MLE for $\theta $ parameter in beta distribution $B(\theta, 2)$:
$$f(x, \theta) = \theta(\theta+1)x^{\theta-1}(1-x) {1}_{\{(0, 1)\}}(x)$$
My strategy will be a standard one - calculate derivative and look for zero of a function.
$$L(X, \theta) = \theta^n (\theta+1)^n (x_1x_2...x_n)^{\theta-1} (1-x_1)(1-x_2)...(1-x_n){1}_{(0,1)}(x_1){1}_{(0,1)}(x_2)...{1}_{(0,1)}(x_n)$$
$$\ln L(X, \theta) = \ln(\theta^n) + \ln(\theta+1)^n + \ln(x_1...x_n)^{\theta-1}+\ln((1-x_1)...(1-x_n)) + \ln({1}_{(0,1)}(x_1)...,{1}_{(0,1)}(x_n))$$
$$\ln L(X,\theta) = n\ln(\theta) + (\theta - 1)\ln(x1,...,x_n) + \ln((1-x_1)...(1-x_n)) + \ln({1}_{(0,1)}(x_1)...,{1}_{(0,1)}(x_n))$$
$$\frac{\partial \ln L(X, \theta)}{\partial \theta} = \frac{n}{\theta} + \frac{n}{\theta} + \ln(x_1,...x_n)$$
$$\frac{\partial \ln L(X, \theta)}{\partial \theta} = 0 \Leftrightarrow \frac{n}{\theta} + \frac{n}{\theta + 1} + \ln(x1 \cdot x2 \cdot ... \cdot x_n)= 0$$
$$(\theta + 1)n + \theta n + \theta(\theta + 1)\ln(x1 \cdot x2 \cdot ... \cdot x_n) = 0$$
$$\theta^2 \ln(x1 \cdot x2 \cdot ... \cdot x_n) + \theta (\ln(x1 \cdot x2 \cdot ... \cdot x_n) + 2n) + n =0$$
$$\theta^2 \cdot \frac{\ln(x1 \cdot x2 \cdot ... \cdot x_n)}{n} + \theta (\frac{\ln(x1 \cdot x2 \cdot ... \cdot x_n)}{n} + 2) + 1 = 0$$
For simplicity denote  $a_n:= \frac{\ln(x1 \cdot x2 \cdot ... \cdot x_n)}{n}$
$$\theta^2a_n + \theta(a_n + 2) + 1 = 0$$
$$\Delta = (a_n + 2)^2 - 4a_n = a_n^2 + 4 \Leftrightarrow \sqrt{\Delta} = \sqrt{a_n^2 + 4}$$
We have two solutions:
$$\theta_1 = \frac{-a_n - 2 - \sqrt{a_n^2 + 4}}{2a_n}$$
$$\theta_2 = \frac{-a_n - 2 + \sqrt{a_n^2 + 4}}{2a_n}$$
I know that $a_n < 0$ since $X_1, X_2,...,X_n \in [0, 1] \Rightarrow \frac 1 n \sum_{i = 1}^n \ln(X_i) < 0$.
I wanted to show somehow that one of them is meaningless - for example by showing that $\theta_1$ or $\theta_2$ only take negative values but I don't think its true.
My question is: Do I have two maximum likelihood estimators or I should get rid of one?

For $a_n < 0$, $$\frac{-a_n - 2 \pm \sqrt{a_n^2 + 4}}{2a_n} = -\frac{1}{2} - \frac{1}{a_n} \mp \frac{1}{2}\sqrt{1 + \frac{4}{a_n^2}}.$$  Notice the reversal in sign due to the fact that $a_n < 0$ implies $a_n = -\sqrt{a_n^2}$.  Therefore, as $a_n \to -\infty$, we have $\theta_1 \to -\frac{1}{2} + \frac{1}{2} = 0$, but $\theta_2 \to -\frac{1}{2} - \frac{1}{2} = -1$.  This would suggest $\theta_1$ is the correct root.
