I once spent far too long getting nowhere with this.
Is there a way of finding the real roots of $ax^k-bx^{k-1}+b-a=0$ where $a, b, k\in \mathbb N$ and $b\gt a$ and $k\gt 1$?
I know that there is no general formula for solving polynomials of degree greater than 4, but with so few $x$s I thought it might be possible. Note that $x=1$ is always a solution.
Because of the dearth of $x$s the stationary points are easy to find, and I know a solution exists between $x=\frac{b(k-1)}{ak}$ and $x=\frac{b}{a}$.

  • $\begingroup$ I assume you have done quite some work already. What do you know about the derivative? Where do you expect to find other real roots, besides between those two values? $\endgroup$
    – TMM
    Commented Sep 25, 2011 at 19:36
  • $\begingroup$ For odd k there is a maximum at x=0 and a minimum at $\frac{b(k-1)}{ak}$. There are the two solutions mentioned in my question and a negative one. For even k x=0 becomes a point of inflection. There are only the two solutions. $\endgroup$ Commented Sep 25, 2011 at 19:50
  • 3
    $\begingroup$ Notice that since $x=1$ is always a solution, remaining real roots satisfy $a x^{k-1} = (b-a) \sum_{n=0}^{k-1} x^n$, which does not look as innocent as you original equation. For example, for $k=6$, $a=1$ and $b=2$, the equation $x^5 = 1+x+x^2+x^3+x^4$ admits no solution in radicals. $\endgroup$
    – Sasha
    Commented Sep 25, 2011 at 20:37
  • $\begingroup$ Ah Sasha. You commented while I was typing up. I'm too slow. $\endgroup$
    – jspecter
    Commented Sep 25, 2011 at 21:03

2 Answers 2


Let's examine the specific case where $a = 1,$ $b = 2$ and $n = 6$ i.e. consider the polynomial $$f(X) = X^6 - 2X^5 + 1.$$

As you've pointed out, $1$ is a root of $f$ and hence $X-1$ divides $f$ in $\mathbb{Q}[X].$ In fact,

$$f(X) = (X -1)(X^5 -X^4 -X^3 -X^2 - X - 1).$$

So let's instead consider the polynomial $$g(X) = X^5 -X^4 -X^3 -X^2 - X - 1.$$ We claim $g(X)$ is not solvable by radicals.

First observe that $g(X)$ is irreducible over $\mathbb{Q}$ as it's reduction modulo $5$ is irreducible over $\mathbb{F}_5.$

Let $L$ be the splitting field of $g$ over $\mathbb{Q}$ and $G = Gal (L/\mathbb{Q}).$ There is a faithful representation of $G \rightarrow S_5$ given by the action of $G$ on the roots of $g,$ identify $G$ with it's image under this representation.

As $L$ is the splitting field of a irreducible fifth degree polynomial, we have $5| |G|$. And so $G$ contains an element of order $5.$ As the only elements in $S_5$ of order $5$ are $5$ cylces, we obtain $G$ contains a $5$-cycle.$

We claim that complex conjugation restricted to $G$ has a cycle decomposition equal to the product of two $2$-cycles. Note that this is equivalent to showing $g$ has one real root. So we show the latter.


$$f'(X) = 6X^5 - 10X^4 = 2X^4(3X^4-5).$$

has $2$ real roots. It follows $f$ has at most $3$ real roots. So $g$ has at most $2$ real roots. As complex roots of a polynomial over $\mathbb{R}$ necessarily come in conjugate pairs and every odd degree polynomial over $\mathbb{R}$ has a real root, it must be the case that $g$ has exactly one real root.

So $G$ contains a five cycle and an element with a cycle decomposition equal to the product of two $2$-cycles. It follows $G$ contains $A_5$ and $G$ is not solvable. Consequently, $g$ will not be solvable by radicals.


Twelve years too late !

I suppose that you look for the largest $x$ solution of $$ax^k-bx^{k-1}+b-a=0$$

Let $b= c a$ with $c>1$ and search for the zero of function $$f(x)=x^k- c\, x^{k-1}+(c-1)$$

As you noticed $f(c)=c-1$ which is positive. So use one single iteration of any Newton-like method of order $n$ with $x_0=c$.

You would obtain $$x_1^{(2)}=c-(c-1) c^{1-k}$$ $$x_1^{(3)}=c-\frac{(c-1) c}{c^k-c k+c+k-1}$$

For the spcific case used by @jspecter $(c=2,k=6)$, we should obtain $$x_1^{(2)}=\frac{63}{32}=1.96875 \quad \text{and} \quad x_1^{(3)}=\frac{116}{59}=1.96610$$ while the solution is $1.96595$.

Now, using higher orders, we generate the sequence $$\left\{\frac{63}{32},\frac{116}{59},\frac{3407}{1733},\frac{20 0108}{101787},\frac{23506334}{11956741},\frac{23506334}{119 56741},\frac{40544849025}{20623558784}\right\}$$ The last one is in an absolute error of $1.167\times 10^{-11}$.

Be sure that we can do much better.

  • $\begingroup$ (+1) It's never too late! Thank you for an alternate approach. $\endgroup$ Commented Aug 21, 2023 at 14:15

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