Determine roots of a polynomial with variable exponent I need to know the nature of the roots of the equation $$ x(x+a)^b -1 = 0 $$ when changing a and b, where $ a,b  $ are natural numbers, I've looked around on the web but I was unable to find how to do it.
 A: Here is another way, without the calculus, but still giving insight.  
$x(x+a)^b-1$ is a polynomial with exactly $1$ sign change, so by Descartes' rule of signs, there is always exactly one positive root which clearly is in $(0, 1)$.  
Let us now seek negative roots. With $y =x+a$, the polynomial is $(y-a)y^b-1 = y^{b+1}-ay^b-1$, so with only one sign change, we know there are no roots for $x \in (-a, 0)$.  
Further setting $y \to -z$, we get $(-1)^{b+1}(z^{b+1}+az^b)-1$, which has no sign change if $b$ is even, and exactly $1$ sign change if $b$ is odd. 
Putting it together, we have for even $b$ exactly one real root in $(0, 1)$ and for odd $b$, one root in $(0, 1)$ and another $< -a$. 
A: Let $P_{a,b}=x(x+a)^b-1$.
If $a=0$ the problem is easily solved. So let $a\neq 0$.
$P'_{a,b}(x)=(x+a)^b+bx(x+a)^{b-1}=(x+a)^{b-1}((1+b)x+a)$
So, now we must make a couple of cases:

*

*if $b$ is odd, then $P'_{a,b}(x)>0\Longleftrightarrow x>-\frac{a}{1+b}\ $ so $P_{a,b}$ has at most $2$ real roots, depending on the sign of the global minimum $P_{a,b}\left(-\frac{a}{1+b}\right)=-\frac{a^{b+1}b^b}{(1+b)^{b+1}}-1<0$
So $P_{a,b}$ has exactly $2$ real roots if $b$ is odd.


*if $b$ is even, then $P'_{a,b}(x)>0\Longleftrightarrow x<-a\vee x>-\frac{a}{1+b}$; this tells us that $P_{a,b}$ has a local maximum in $x=-a$ and a local minimum in $x=-\frac{a}{1+b}$. Notice, though, that $P_{a,b}(-a)<0$. So there can be only one real root.
A: If $a=0$, there are trivially one or two real roots, depending on $b$ being even/odd.
Assume now that $a>0$ and $b$ is odd. $f(x)=x(x+a)^b$ has just two real roots, $x=0$ and $x=-a$. By computing $f'(x)$, it is easy to check that $f(x)$ has an absolute minimum $\xi\in(-a,0)$, $f(\xi)<0$, $f(x)$ is decreasing over $(-\infty,\xi)$ and increasing over $(\xi,+\infty)$, so $f(x)=1$ has just two real roots, one root close to zero and the other one close to $-a-1$.
In the same way we may check that if $a>0$ and $b$ is even, $f(x)=x(x+a)^b$ is non-positive over $\mathbb{R}^-$ and increasing over $\mathbb{R}^+$, so $f(x)=1$ in just one point, close to zero.
A: There is a new special and multivalued function called Lambert-Tsallis, where you can find details about at "Physica A: Statistical Mechanics and its Applications 525, 164-170", or to get in contact with a friend and coworker R. V. Ramos.
The solution to your problem is given by
$x= \frac{a}{b}\cdot W_q( \frac{b}{a^{1+b}} )$ with $q= \frac{b-1}{b}$.
For the case where a and b are naturals, this solution:

*

*It will have $b+1$ solution.

*When $b$ is even, there is only one positive solution the other are complex (and is conjugate)

*When $b$ is odd, there will be two real solution with oposite signals and the others will appear in complex conjugated pairs.

One interesting solution arises when $a=b=1$ because $x$ will value  the negative of both golden numbers, because $W_0(1)$.
