How to solve a quartic equation given that the roots are all part of an arithmetic sequence?

Question

If I have a quartic, and the roots are all members of an arithmetic progression (i.e. the roots are $\alpha$, $\alpha +k$,$\alpha +2k$ and $\alpha +3k$), is there a way to solve for $x$?

e.g. I have the quartic: $x^4-16x^3+86x^2-176x+105=0$.

Given that it is written in the form: $ax^4+bx^3+cx^2+dx+e$, I have worked out that:

$\frac{-b}{a}=4\alpha+6k$

$\frac{c}{a}=6\alpha^2+18\alpha k+11k^2$

$\frac{-d}{a}=4\alpha^3+18\alpha^2k+22\alpha k^2 +6k^3$

$\frac{e}{a}=\alpha^4 +6\alpha^3k+11\alpha^2k^2+6\alpha k^3 $

Answer 1

When expanding, the constant term is $$\alpha (\alpha+k)(\alpha+2k)(\alpha+3k)=$$

$$105=1×3×5×7$$ $$=1×(1+2)×(1+2×2)×(1+3×2) $$ thus

$$\alpha=1$$ and $$k=2$$

Answer 2

You should be confident with your method, this was almost finished.

$4\alpha+6k=\frac{-b}{a}=16\iff \alpha=4-\frac 32k$

$6\alpha^2+18\alpha k+11k^2=\frac{c}{a}=86\iff 96-\frac 52k^2=86\iff k^2=4\iff k=\pm 2$

So we get $a=7,k=-2$ or $a=1,k=2$ both giving the list $1,3,5,7$.

And then you can verify that $(x-1)(x-3)(x-5)(x-7)$ matches the initial polynomial.

Answer 3

Because the roots are defined to be in arithmetic progression, the reduced quartic, obtained with the substitution:

$$x = y - \dfrac{b}{4a}$$

is an even symmetric function of the form:

$$ a'y^4 + c'y^2 + e' = 0$$

with roots at $\left(-\dfrac{3}{2}k, -\dfrac{1}{2}k, \dfrac{1}{2}k, \dfrac{3}{2}k\right)$ to which one applies the quadratic formula in the process of finding $k$.

The non-zero coefficients of the reduced quartic are:

$a' = a$

$c' = 6a\left(\dfrac{b}{4a}\right)^2-3b\left(\dfrac{b}{4a}\right)+c$

$e' = a\left(\dfrac{b}{4a}\right)^4-b\left(\dfrac{b}{4a}\right)^3+c\left(\dfrac{b}{4a}\right)^2-d\left(\dfrac{b}{4a}\right)+e$

So the solution for $k$ is:

$$k = 2 \sqrt{\dfrac{-c' - \sqrt{(c')^2 -4a'e'}}{2a'}}$$

or equivalently:

$$k = \dfrac{2}{3} \sqrt{\dfrac{-c' + \sqrt{(c')^2 -4a'e'}}{2a'}}$$

or, using those two expressions for $k$ together, because of the special nature of the roots being spaced by $k$, it appears $k$ can also be expressed as:

$$k = \sqrt{\dfrac{-2c'}{5a'}}$$ The solution for $\alpha$ is:

$$\alpha = -\dfrac{b}{4a} - \dfrac{3}{2}k$$

Here's a function written in Octave/MatLab for testing the above:

function [alpha, k, x] = symmetric_quartic(p)
%%
%% [ALPHA, K, X] = SYMMETRIC_QUARTIC(P)
%%
%% Solve a special quartic polynomial P, whose roots, X, are known
%% to be real and in an arithmetic progession:
%% ALPHA, ALPHA + K, ALPHA + 2K, ALPHA + 3K.
%%

a = p(1);
b = p(2);
c = p(3);
d = p(4);
e = p(5);

x_N = b/(4*a);
a_r = a;
b_r = 0;
c_r = 6*a*x_N^2 - 3*b*x_N + c;
d_r = -4*a*x_N^3 + 3*b*x_N^2 - 2*c*x_N + d;
if (d_r != 0)
  printf('Error: reduced quartic is not an even function as expected');
  alpha = 0;
  k = 0;
  x = [];
  return
endif
e_r = a*x_N^4 - b*x_N^3 + c*x_N^2 - d*x_N + e;

k = 2*sqrt((-c_r - sqrt(c_r^2 - 4*a_r*e_r))/(2*a_r));

alpha = -x_N - 3/2*k;

x = [alpha; alpha+k; alpha+2*k; alpha+3*k];

return