Finding method of moments estimate for density function $f(x|\alpha) = \frac {\Gamma(2\alpha)} {\Gamma(\alpha)^2}[x(1-x)]^{\alpha - 1}$ 
Suppose that $X_1, X_2, ..., X_n$ are i.i.d random variables on the interval $[0,1]$ with the density function
$$
f(x|\alpha) = \frac {\Gamma(2\alpha)} {\Gamma(\alpha)^2}[x(1-x)]^{\alpha - 1}
$$
where $\alpha > 0$ is a parameter to be estimated from the sample. It can be shown that
\begin{align}
E(X) &= \frac 1 2 \\
\text{Var}(X) &= \frac 1 {4(2\alpha +1)}
\end{align}
How can the method of moments be used to estimate $\alpha$?

My attempt
It is clear that the first moment of $X$ is $\mu_1 = E(X) = \frac 1 2$.
The second moment of $X$ is given by
\begin{align}
\mu_2 &= E(X^2) \\
&= \text{Var}(X) + (E(X))^2 \\
&= \frac 1 {4(2\alpha + 1)} + \frac 1 4 \\
&= \frac {\alpha + 1} {2(2\alpha + 1)}
\end{align}
Thus we have the relation
$$
\alpha = \frac {1 - 2\mu_2} {4\mu_2 - 1}
$$
Using the method of moments, we obtain
\begin{align}
\hat{\alpha} &= \frac {1 - 2\hat{\mu_2}} {4\hat{\mu_2} - 1} \\
&= \frac {1 - \frac 2 n \sum_{i=1}^n X_i^2} {\frac 4 n \sum_{i=1}^n X_i^2 - 1} \\
&= \frac {n - 2 \sum_{i=1}^n X_i^2} {4 \sum_{i=1}^n X_i^2 - n}
\end{align}
Solution provided
$$
\hat{\alpha} = \frac n {8 \sum_{i=1}^n X_i^2 - 2n} - \frac 1 2
$$
Did I apply the method of moments correctly for this question? I can't seem to obtain the form as suggested in the sample solution provided. Any advice would be greatly appreciated!
 A: You are right. In fact
\begin{align}
\frac n {8 \sum_{i=1}^n X_i^2 - 2n} - \frac 1 2&= \frac{2n - 8 \sum_{i=1}^n X_i^2 + 2n}{16 \sum_{i=1}^n X_i^2 - 4n}\\
&= \frac{4n - 8 \sum_{i=1}^n X_i^2}{16 \sum_{i=1}^n X_i^2 - 4n}\\
&=\frac {n - 2 \sum_{i=1}^n X_i^2} {4 \sum_{i=1}^n X_i^2 - n}\,.
\end{align}
The last equality is just a simplification by $4$.
A: Your answer is equivalent to the solution provided, so there is no problem there.  It is also worth noting that there is an alternative MOM estimator that is obtained by equating the sample variance with the variance expression (as opposed to using the second raw moment).  Since the mean is known to be $\mathbb{E}(X)=\tfrac{1}{2}$, the sample variance incorporating this information is:
$$S_*^2 = \frac{1}{n} \sum_{i=1}^n (X_i-\tfrac{1}{2})^2 = \frac{1}{n} \sum_{i=1}^n X_i^2 - \frac{1}{n} \sum_{i=1}^n X_i + \frac{1}{4}.$$
Estimating via the variance (instead of the second raw moment) gives an alternative MOM estimator which satisfies $S_*^2 = 1/[4 (2 \hat{\alpha}_* + 1)]$, which gives the explicit form:
$$\hat{\alpha}_* = \frac{n}{8 \sum x_i^2 - 4\sum x_i + 2n} - \frac{1}{2}.$$
These estimators are quite similar, but they will have slightly different performance.  It would be worth running some simulations on these two estimators to find their MSE over the parameter range.
