Q:

Showing a system of equations having two solutions in $\mathbb{R}^2$

Consider the system of equations in $\mathbb{R}^2$ \begin{align*}
\xi^2 + \eta^2 &= 4\\
e^{\xi} + \eta &= 1
\end{align*}
  Show that the system has two solutions in $\mathbb{R}^2$ has two solutions in  $\mathbb{R}^2$, one with $\xi > 0$ and $\eta < 0$ and one with $\xi < 0$ and $\eta > 0$.

Attempted solution - The two equations are a circle of radius $2$ center at the origin and exponential opening down and to the left. If we rewrite equation $2$ as $\eta = 1 - e^{\xi}$ and plug in to equation $1$ we get $$\xi^2 - 2e^{\xi} + e^{2\xi} = 3$$
Using Wolfram we find two solution: $\xi = - 1.81626$ and $\xi = 1.00417$. Hence for $\xi = 1.00417$ we have $\eta = -1.7296$ and when $\xi = -1.81626$ we have $\eta = .83737$ satisfying the condition in the problem.
However, I am curious if there is a way of obtaining an interval to where the solution exists given the two conditions without using software of some kind. Any suggestions is greatly appreciated.

A:

Note that they only want you to show that there are two solutions, not solve for the solutions exactly. I would probably try to show that there are at least two solutions using the intermediate value theorem. Then that there cannot be more from comparing derivatives.

A:

Let $f(\xi ) =\xi ^2-2e^{\xi}+e^{2\xi}-3$.  Note that $$f(-2)=1-\frac{2}{e^2}+\frac{1}{e^4}>0 $$ and $$f(\ln(2))=\ln(2)^2-3<0$$ and $$f(\ln(3))=\ln(3)^2>0$$  By the Intermediate Value Theorem, you have at least two roots.  

