Is method of Moments on a Uniform distribution $U(a,b)$ unbiased?

Let $(X_i)_{i=1}^n$ forms a random sample from the uniform distribution $U(a,b)$. The parameters $a$ and $b$ are both unknown. When we use method of moments we have $$\hat{a} = \hat{\mu}_1 -\sqrt{3(\hat{\mu}_2 - \hat{\mu}_1^2)}$$ $$\hat{b} = \hat{\mu}_1 +\sqrt{3(\hat{\mu}_2 - \hat{\mu}_1^2)}$$ where $$\hat{\mu}_1 = \frac{\sum_{i=1}^n X_i}{n}$$ $$\hat{\mu}_2 = \frac{\sum_{i=1}^n X_i^2}{n}$$

I want to know is this estimator is unbiased? I think we must verify $\mathbf{E}(\hat{a}) =a$ and $\mathbf{E}(\hat{b}) = b$. But I don't know how to calculate $\mathbf{E}(\sqrt{\hat{\mu}_2 - \hat{\mu}_1^2})$(it is really difficult). I really need some help.

Jensen's inequality gives that both estimators are biased. Indeed,$\sqrt{x}$ is (strictly) concave function, and for any non-negative r.v. $X$ with non-degenerate distribution, $$ \mathbb E[\sqrt{X}] < \sqrt{\mathbb E[X]}. $$ Therefore $$ \mathbb E[\hat{a}] = \mathbb E[\hat{\mu}_1] -\mathbb E\left[\sqrt{3(\hat{\mu}_2 - \hat{\mu}_1^2)}\right] > \frac{a+b}{2} - \sqrt{3\mathbb E[\hat{\mu}_2 - \hat{\mu}_1^2]}. \tag{1}\label{eq1} $$ Next, $$ \mathbb E[\hat{\mu}_2 - \hat{\mu}_1^2]=\frac{n-1}{n}\text{Var}[X_1]=\frac{n-1}{n}\frac{(b-a)^2}{12}<\frac{(b-a)^2}{12}. $$ Continuing the inequality \eqref{eq1}, get $$ \mathbb E[\hat{a}] > \frac{a+b}{2} - \sqrt{\frac{(b-a)^2}{4}} = a. $$ The same way you can get $\mathbb E[\hat{b}]<b$.