$e^x= x^2+x$ has no roots for $x>0$ I need to prove using elementary calculus that:
$$e^x=x^2+x$$
has no roots for $x>0$.
I could easily observe it graphically but how can I prove it.
Please suggest.

We have $e^x > 1+x+x^2/2+x^3/6$, so we show that the right side is bigger than $x^2+x$ for positive $x$.  Let
$$f(x) = 1+x+x^2/2+x^3/6-(x^2+x) = x^3/6-x^2/2+1.$$
Find the minimum value of $f(x)$ using usual calculus methods:
$$f'(x) = x^2/2-x = 0$$
gives one positive critical point $x=2$.  $f''(x) = x-1$ and $f''(2) >0$ so the critical point is a minimum.  Since $f(2) = 1/3>0$ the minimum value of $f$ on $x>0$ is positive.
