Showing that $x^n-ax^{n-1}-bx-1$, for non-negative integers $n\geq3$, $a\geq1$, $b$, has no double roots 
Let $f(x)=x^n-ax^{n-1}-bx-1$ be a polynomial, where $n, a, b$ are non-negative integers, with $n\geq 3$ and $a\geq 1$.
I would like to prove that this polynomial does not have any double root. (This is supported with a Mathematica routine for $(n,a,b)\in [3,10]\times [1,50]\times [0,50]$).

In fact, I was able to show that the only possible double root must be a real negative number $x_0$ (actually a quadratic irrational number) which can happen only for the case $n$ odd.
Of course, I tried to study the system $f(x)=f'(x)=0$ which lead to the quadratic equation
$$
b(n-1)x^2+(ab-a(n-1)+n)x-a(n-1)=0.
$$
(Indeed from $f(x)=0$ and $f'(x)=0$, we obtain $x^{n-1}(x-a)=bx+1$ and $x^{n-2}(nx-(n-1)a)=b$. It is easy to see that $f(x)$ does not have rational roots - rational root theorem - so we can combine
$$
x^{n-1}=\frac{bx+1}{x-a}\ \mbox{and}\ x^{n-2}=\frac{b}{nx-(n-1)a}
$$
to obtain the previous quadratic relation).
Also it is possible to prove that such a double root must satisfy a relation like
$$
ax^{n-1}+b(n-1)x+n=0.
$$
So, I tried to use some arithmetic behaviour of power of quadratic irrationals of the form $x_0=(r+s\sqrt{\xi})$, where $r,s\in \mathbb{Q}$ and $\xi$ is a non-square positive integer. However, without success.
Of course, it is possible to impose conditions on $a$ and $b$ for which we do not have double roots (as for example, $b>a+2$ forces $f(-1)>0$ and so we have exactly two negative roots, counted with multiplicity. Here we used Descartes' sign rule).
Thus, I would like to ask you guys for some suggestion. Thanks!
 A: Let $f\in\mathbb{Z}[x]$ be given by
$$
f=x^n-ax^{n-1}-bx-1$$
where $n\ge 3,a\ge 1,b\ge 0$

Claim:$\;f$ has no repeated roots.

Proof:

Suppose otherwise.

Our goal is to derive a contradiction.

First note that if $f$ has a rational root $r$, then

*

*$r=-1$.$\\[4pt]$

*$n$ is odd.$\\[4pt]$

*$b=a+2$.$\\[4pt]$

*$f'(-1)=(a+1)(n-2) > 0$.

hence no repeated root of $f$ is rational.

Let $w$ be a repeated root of $f$.

As you showed (but with a minor correction to your result), $w$ must satisfy the equation
$$
b(n-1)w^2-\bigl(ab(n-2)-n\bigr)w-a(n-1)=0
$$
so we can't have $b=0$, else $w$ would be rational.

Hence $w$ is a root of the quadratic polynomial
$$
p=x^2-\left(\frac{ab(n-2)-n}{b(n-1)}\right)x-\frac{a}{b}
$$
which must be irreducible in $\mathbb{Q}[x]$ (since $w$ is not rational).

Since $p$ is irreducible in $\mathbb{Q}[x]$ and $w$ is a common root of $p$ and $f$, it follows that $p$ is a factor of $f$ in $\mathbb{Q}[x]$, hence we can write $f=pg$ for some $g\in\mathbb{Q}[x]$. 

But $w$ is a simple root of $p$ (since $p$ is irreducible in $\mathbb{Q}[x]$), and by assumption, $w$ is a repeated root of $f$, hence from $f=pg$, we get that $w$ is a root of $g$, so $p$ is a factor of $g$ in $\mathbb{Q}[x]$.

It follows that $p^2$ is a factor of $f$ in $\mathbb{Q}[x]$.

Note that the discriminant of $p$ is positive, so the roots of $p$ are real, and the product of the roots of $p$ is negative, so one of the roots of $p$ is negative, and the other is positive.

But the roots of $p$ are repeated roots of $f$, contradiction, since by Descartes' rule of signs, $f$ can't have a repeated positive root.
