# Show $a e^x=1+x+\frac{x^2}{2}$ has exactly one real root

I'm struggling with the following problem:

Show that the equation $a e^x=1+x+x^2$, where $a$ is a positive constant, has exactly one real root.

Looking at the graphs of both $ae^x$ and $1+x+x^2$, they will intersect on the left half of the plane when $a>1$ and on the right half when $a<1$ (because a makes $e^x$ grow faster or slower). So if I could find a point $x_1$ where $a e^{x_1}-(1+x_1+{x_1}^{2})<0$ and a point $x_2$ where $a e^{x_2}-(1+x_2+{x_2}^2)>0$, by the intermediate value theorem, the equation would have a real root. Also,since the derivative of $a e^x-(1+x+x^2)$ is always positive, I could colclude that there can't exist another real root.

So my question is, how can I find $x_1$ and $x_2$? I can't figure out how to express $x$ in terms of $a$.

EDIT: I'm really sorry, I wrote the equation wrong everywhere. I should read $ae^x=1+x+\frac{x^2}{2}$ instead of $ae^x=1+x+x^2$. Changed to reflect that.

EDIT2: The question appears as it was originaly written. I'll make new question with the right equation. Sorry if I bothered anyone.

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But do we really know the derivative of $ae^x -(1+x+x^2)$ is always positive? I get a derivative of this to be $ae^x - (1+2x)$, and it isn't obvious to me that even assuming $a \gt 1$ suffices to make the derivative positive everywhere. But if you knew that, and could find points above and below the x-axis, yes it would prove there's exactly one real root. – hardmath Jan 5 '13 at 17:28
You really shouldn't change your question when you already have received answers to the original one. (It's a bit disrespectful of the work that was put into answering the originial question...) It's better to post a new question. – mrf Jan 5 '13 at 17:55
Sorry, It was a silly mistake from my part not to check that I wrote the correct data, I'll post a new one. I swear I didn't mean to be disrespectful. – user55286 Jan 5 '13 at 17:58
@mrf It completely changes the answer, but it changes none of the methods used to solve it. – Steven Stadnicki Jan 5 '13 at 18:00
See meta.math.stackexchange.com/questions/2304/… and decide for yourself. My answer will have to be deleted if you stick with the new question, since it doesn't apply to the updated version. – mrf Jan 5 '13 at 18:04

This is not true. Let $f(x) = ae^x - 1 - x- x^2$. If $a = 1$, then $f(0) = 0$, but it's easy to check that $f(x)$ is negative for small positive values of $x$ and positive for large values of $x$, so $f$ has one more zero.

(Numerically, the second zero is located at $x \approx 1.793282133$.)

Added: In fact, with more work you can show that the equation has three roots for $1 < a < 3e^{-1} \approx 1.103638324$ (and two roots, one of which is double for $a=1$ and $a=3e^{-1}$).

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I believe some values of $a$ give three solutions. e.g., $a=1.01$ (from examining a plot). – David Mitra Jan 5 '13 at 17:45
@DavidMitra You're right of course, I typed that too quickly, the double root at $x=0$ when $a=1$ splits into two single roots for $a > 1$. – mrf Jan 5 '13 at 17:47

To show the existence of $x_1$, I would take a look at the behaviour of the $ae^x-1-x-x^2$, when x goes to $-\infty$. The limes should be $-\infty$? - Maybe use L'Hopital.

Therefore the function should be negative anywhere. The same procedure should lead you to a point $x_2$, where the function is positive.

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Did you mean limes or limit? – Amzoti Jan 5 '13 at 23:18