I'm unsure how to prove the following equation
$(1+x^2)e^x = k$
where $k$ is a constant, has one real root where $k>0$ and no real roots if $k<0$.
So far I know the graph and derivative of $(1+x^2)e^x = 0$, as well as having attempted to use the discriminant for the above question with $a =e^x$, $b=0$ and $c = e^x-k$ however I don't seem to get an appropriate answer. Any help would be massively appreciated.
On
If $ k\le 0$, there will be no root because
$$(\forall x\in\Bbb R)\;\; (1+x^2)e^x>0$$
assume $ k>0$ and put $$f(x)=(1+x^2)e^x$$
$ f $ is differentiable at $ \Bbb R$ and
$$(\forall x\in \Bbb R)\; f'(x)=(1+x)^2e^x>0$$ $f $ is continuous and strictly increasing at $ \Bbb R $, thus $ f $ is a bijection from $ \Bbb R$ to
$$(\lim_{-\infty}f,\lim_{+\infty}f)=(0,+\infty)$$
So $$(\forall k>0) \;\; (\exists\; ! \;x\in \Bbb R)\;:\;f(x)=k$$
It has at most one real root. Let $f(x)=e^x(1+x^2)$. Notice that $$\frac{\mathrm{d}}{\mathrm{d}x}f(x)=e^x(1+2x+x^2)\geq 0 ~ \forall x\in \Bbb{R}.$$ I.e, the function is non-decreasing. Therefore it can only attain the value of $k$ at one point, assuming $f$ is never "completely flat".