$e^x= x^2+x$ has no roots for $x>0$

Question

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.

Answer 1

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.

Answer 2

Fix a positive integer $n$ and define $$ f_n(x):=e^x-1-x-x^2/2-\cdots-x^n/n!. $$ Then $f^\prime_n(x)=e^x-1-x-\cdots-\frac{x^{n-1}}{(n-1)!}$, which is positive (by induction) on $(0,\infty)$. Hence $f_n$ in strictly increasing on $(0,\infty)$. In particular, this gives the known inequality $$ \forall x>0, \quad e^x>1+x+\cdots+\frac{x^n}{n}. $$

Suppose that $z$ satisfies $e^z=z^2+z$. Then $$ z^2+z=e^z=f_n(z)+1+z+\cdots+\frac{z^{n}}{n!}>f(0)+1+z+\cdots+\frac{z^n}{n!}=1+z+\cdots+\frac{z^n}{n!} $$ for all $n$. Now pick $n=1$ and you know that $z>1$. Set $y:=z-1 \in (0,\infty)$, so that $$ y^2+3y+2=z^2+z=e^z=e\cdot e^y. $$ Using the above inequality with $n=2$ we obtain $$ e\cdot e^y> e\left(1+y+\frac{y^2}{2}\right)>y^2+3y+2, $$ where the last inequality is equivalent to $$ y^2+2y\left(1-\frac{1}{e-2}\right)+2>0, $$ or, equivalently, $$ \left(y+1-\frac{1}{e-2}\right)^2+\left(2-\left(1-\frac{1}{e-2}\right)^2\right)>0 $$ However, the first square is nonnegative and the second one is positive.