SOLUTION 9: We are given the equation
$$x e^x = x^2-1 \ \ \ \ \longrightarrow \ \ \ \ x e^x - x^2+1=0$$
Let function
$$f(x)= xe^x - x^2+1 \ \ \ \ and \ choose \ \ \ \ m=0$$
This function is continuous for all values of $x$ since it is the SUM of continuous functions. We have the continuous polynomial function $1-x^2$. And the function $xe^x$ is continuous for all values of $x$ since it is the PRODUCT of $x$, a polynomial, and $e^x$, a well-known continuous function. We now need to search for an appropriate interval satisfying the assumptions of the Intermediate Value Theorem. By trial and error, we have that
$$f(0)= (0)e^0 - (0)^2+1= 1 > 0 \ \ \ \ and \ \ \ \ f(-2)= (-2)e^{-2} - (-2)^2+1= \displaystyle{ -2 \over e^2 }-3 \approx -3.27 <0$$
so that $$f(-2) \approx -3.27 < m < 1 = f(0)$$
i.e., $m=0$ is between $f(-2)$ and $f(0)$.

The assumptions of the Intermediate Value Theorem have now been met, so we can conclude that there is some number $c$ in the interval $[-2, 0]$ which satisfies $$f(c)=m$$ i.e., $$ce^c - c^2+1 =0$$ and the equation is solvable.