Let $x_t=(b_0\epsilon_t+b_1\epsilon_{t-1})$ be a MA(1) process. Assume that the autocovariance function $\gamma(k)$ is given. Use $\gamma(k)$ to determine the parameters $(b_1,b_1,\sigma^2)$. Without loss of generality $b_0=1$.
First of all, why can we choose $b_0=1$?
I tried (using $b_0=1)$
$\gamma(k)=\sigma^2\sum_{j=0}^{1}b_{j+k}b_j=\begin{cases}\sigma^2(1+b_1^2)&&k=0\\\sigma^2b_1&&|k|=1\\0&&|k|>1\end{cases}$
And for the autocorrelation function we get:
$\rho(k)=\frac{\gamma(k)}{\gamma(0)}=\begin{cases}1&&k=0\\\frac{b_1}{b_1^2+1 }&&|k|=1\\0 &&|k|>1\end{cases}$
So from $\rho(1)$ we receive $$\frac{\gamma(1)}{\gamma(0)}=\frac{b_1}{b_1^2+1}\Leftrightarrow\gamma(1)b_1^2-b_1\gamma(0)+\gamma(1)=0$$
Solving this equation:
$$b_1=\frac{\gamma(0)\pm\sqrt{y(0)^2-4\gamma(1)^2}}{2\gamma(1)^2}, \ \ \ \gamma(1)\neq 0$$
If $\gamma(1)=0$, then $0=\frac{\gamma(1)}{\gamma(0)}=\rho(1)=\frac{b_1}{b_1^2+1}\Rightarrow b_1=0$ but $b_1$ cant be equal to $0$ because then $(x_t)$ wouldn't be a MA(1) process.
And $\sigma^2=\frac{\gamma(1)}{b_1}$
Does that look correct?