**solved** Proof: the following equation has only one real root

Hi. Could somebody show me how to prove that $\displaystyle ae^x=1+x+\frac{x^2}{2}$ has only one real root for $\displaystyle a>0$ ? All I know so far is that this equation has at least one root because $\displaystyle 1+x+\frac{x^2}{2}>0$ for all real x.

-------------

Already solved

Re: Proof: the following equation has only one real root

I got a solution that i thought was cool so im posting anyways. Now if $\displaystyle ae^x = 1 + x + \frac{x^2}{2} $. then $\displaystyle y = ae^x $ and $\displaystyle y = 1 + x + \frac{x^2}{2} $ should be solutions to the differential eqn $\displaystyle y' - y = 0 $ but for $\displaystyle y = 1 + x + \frac{x^2}{2} $, we get $\displaystyle y ' - y = \frac{x^2}{2} =0 $ implies $\displaystyle x = 0 $ so to have a solution to $\displaystyle ae^x = 1 + x + \frac{x^2}{2} $, x = 0 on the right side, so you get $\displaystyle ae^x = 1 $ which for a > 1, has only 1 real root, namely $\displaystyle x = ln(\frac{1}{a})$