Solve the exponential equation Find the number of roots to $e^x = ax^2$ for all values of $a$. (x is real and so is $a$). I have tried some things but I am stuck.
 A: For $a>0$ the roots can be expressed in terms of Lambert $W$ function:
\begin{align}
a x^2&=\exp(x) \\
x^2\exp(-x)&=\frac{1}{a} \\
x\exp(-\frac{x}{2})&=\pm\frac{1}{\sqrt{a}} \\
-\frac{x}{2}\exp(-\frac{x}{2})&=\mp\frac{1}{2\sqrt{a}} \\
-\frac{x}{2}&=\mathrm{W}\left( \mp\frac{1}{2\sqrt{a}} \right)\\
x&=-2\mathrm{W}\left( \mp\frac{1}{2\sqrt{a}} \right)
\end{align}
So, $\forall a>0$ there is always one negative real root 
$x_1=-2\mathrm{W_0}\left(\frac{1}{2\sqrt{a}} \right)$.
And since the Lambert $W(u)$ function has two real branches
for $-\exp(-1)<u<0$, it would be two more (positive) real roots (three total):
\begin{align}
x_2&=-2\mathrm{W_0}\left(-\frac{1}{2\sqrt{a}} \right), \\
x_3&=-2\mathrm{W_{-1}}\left(-\frac{1}{2\sqrt{a}} \right)
\end{align}
for $a>\exp(2)/4$. 
Also, when $a=\exp(2)/4$ then $x_2=x_3$ and there are two real roots total.
Summarizing, the number $n$ of real roots to $\exp(x) = a x^2$:
\begin{align}
n&=
\begin{cases}
0,\quad a\le0 \\
1,\quad 0<a<\exp(2)/4 \\
2,\quad a=\exp(2)/4 \\
3,\quad a>\exp(2)/4 \\
\end{cases}.
\end{align}

A: I am guessing you are looking for the roots over the whole real line. Note that
$f(x)=e^x-ax^2$ is a continuous function on its domain but more importantly that the domain is a connected set. Does it sound like  Intermediate value theorem?
