# Solve the following system: { <mtable columnalign="left left" rowspacing=".2em" columns

Solve the following system:
$\left\{\begin{array}{l}a{x}_{0}^{2}=\mathrm{exp}\left(-\frac{{x}_{0}^{2}}{4{\sigma }^{2}}\right)+a{r}^{2}\\ \mathrm{exp}\left(-\frac{{x}_{0}^{2}}{4{\sigma }^{2}}\right)+4a{\sigma }^{2}=0\end{array}$
where the varaibles are ${x}_{0},a$. Take ${x}_{0}>0$
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nuvolor8
The solution for $|r|\ge 2\sigma$ is
$\left\{\begin{array}{l}{x}_{0}=\sqrt{{r}^{2}-4{\sigma }^{2}}>0\\ a=-\frac{1}{4{\sigma }^{2}}\mathrm{exp}\left(-\frac{{r}^{2}-4{\sigma }^{2}}{4{\sigma }^{2}}\right)=-\frac{1}{4{\sigma }^{2}}\mathrm{exp}\left(-\frac{{x}_{0}^{2}}{4{\sigma }^{2}}\right).\end{array}$
Since $\mathrm{exp}\left(-\frac{{x}_{0}^{2}}{4{\sigma }^{2}}\right)>0$, the second equation implies that $a<0$. For convenience write
$E=\mathrm{exp}\left(-\frac{{x}_{0}^{2}}{4{\sigma }^{2}}\right).$
The system is easily solvable, e.g. as follows:
$\left\{\begin{array}{l}a{x}_{0}^{2}=E+a{r}^{2}\\ E+4a{\sigma }^{2}=0\end{array}⇔\left\{\begin{array}{l}a{x}_{0}^{2}=-4a{\sigma }^{2}+a{r}^{2}\\ E=-4a{\sigma }^{2}\end{array}⇔\left\{\begin{array}{l}{x}_{0}^{2}=-4{\sigma }^{2}+{r}^{2}\\ a=-\frac{1}{4{\sigma }^{2}}E.\end{array}$