Sum of roots of equation

Question

For k > 0, the set of all values of k for which $$ke^x - x = 0$$ has two distinct roots is (a, b/c), such that b/c is in its lowest form, then what is the value of a + b + c?

Answer 1

If you are after the sum of the two distinct real roots, these can be found. But as Robert Israel notes in his comment, if these roots are written as $(a,b/c)$ then the value for $a + b + c$ cannot be uniquely found as $b/c$ can be written in many different ways.

To find the two real distinct roots, writing the equation as $-x e^{-x} = -k$, on solving for $x$ we have $$x = - \text{W}_\nu (-k).$$ Here $\nu$ denotes the branch of the Lambert W function $\text{W}(x)$.

For two distinct real roots we require $0 < k \leqslant 1/e$. So the two distinct real roots are $-\text{W}_0 (-k)$ and $-\text{W}_{-1} (-k)$ where $\text{W}_0 (x)$ denotes the principal branch of the Lambert W function while $\text{W}_{-1} (x)$ denotes its secondary real branch.

Answer 2

First of all, $k>0$. Then as @vadim123 says, the other side of $k$ is $e^{-1}$. So $0<k<e^{-1}$, but $b/c$ cannot be defined. Maybe $b=1, c=e$, so $a+b+c=e+1$?

If $k<0$, then $f(x)=e^{x-1}$ has negative slope while $g(x)=x$ has positive slope. So they can only have one intersection. Before or after that intersection point, they will start to deviate from each other because of different slopes. $k=0$ is trivial.

For $k>0$, when $k=e^{-1}$, $f(x)=e^{x-1}$ intersects with $g(x)=x$ at $x=1$, and slope $f'(1)=1$, $g'(x)=1$. But $f''(x)=e^{x-1}>0$, meaning $f'(x)>1$ for $x>1$. So $f(x)>g(x)$ for $x>1$. $f'(x)<1, x<1$, so $f(x)>g(x), x<1$. So $x=1$ is the only intersection. When $k>e^{-1}$, $ke^x>f(x)>=g(x)$ meaning there is no intersection. So $0<k<e^{-1}$.