Solve the equation: $e^x=mx^2$ I need to find out the maximum possible number of real roots of the equation:
$$e^x=mx^2$$
where m is a real parameter.
I'm interested in some easy approaches. Moreover, is it possible to solve it without using derivatives at all? Thanks.
 A: After the umpteenth edit I decided to rewrite the solution in a concise way.
Consider the function $f:\mathbb{R} \setminus \{0\}\to \mathbb R$ with $f(x)=\frac{e^x}{x^2}$. Then the number of solutions of the equation $e^x=mx^2$ for a given $m$ is precisely the number of elements in $f^{-1}(m)$. Since $f$ is strictly positive there is no solution for $m\leq 0$. Since the restriction $f:\mathbb R^-\to \mathbb R^+$ is bijective, there is always one negative solution. And since the restriction $f:\mathbb R^+\to \mathbb R^+$ is convex with $f(x)\to \infty$ for $x\to 0^+$ and $x\to \infty$ there are either $0$, $1$ or $2$ positive solutions.
It is easy to see that $f$ restricted to $\mathbb R^+$ is convex using calculus as the second derivative is $\frac{e^x(x^2-4x+6)}{x^4}>0$ but it should also be possible to show this by foot.
A: If $m>0$, there would always be a solution for $x<0$. If m is sufficiently large such that $m*x^2$ overcomes $e^x$ for some $x>0$ then $e^x$ will meet $m*x^2$  again as $e^x$ grows faster than any polynomial giving total of $3$ solutions. If $m$ is such that when $e^x$ meets $m*x^2$ for some $x>0$ and their slopes are equal at that point, then there would be $2$ solutions.If $m>0$ is small, curves would intersect only for some $x<0$ giving only $1$ solution while for $m<0$, they would never intersect giving no real solutions. 
A: It is not completely elementary, but you could try to change variables: $\mathrm{e}^x=t$, so that your equation becomes $$\frac{1}{m}t = \left( \log t \right)^2 .$$
The behavior of the right-hand side should be guessed easily (monotonicity, convexity, and so on). The advantage is that you can probably exploit convexity. But I believe that some differentiation is anyway needed.
A: Provided $m > 0$ there is always a negative solution. We now turn our attention to solutions on $(0,\infty)$.
Put $$g(x) = {e^x\over x^2}$$
for $x > 0$.  Differentiating, you get
$$g'(x) = {(x-2)e^x\over x^3}.$$
If you draw the sign chart for g', you will see it is negative if $x < 2$ and positive if $x > 2$.  Therefore there is a global minimum at  $x = 2$. We conclude that
$${e^x\over x} \ge {e^2\over 4}$$
for $x > 0$.  
So, if $m < e^2/4$, no solution on the $(0,\infty)$ exists.  If $m > e^2/4$, two solutions exist, one to the left of 2 and one to the right.  If $m = e^2/4$, there is one solution.
