Simultaneous Equation Solutions of the Roots and Coefficients of a Polynomial Formulae

Question

It seems to me that equations describing the connection between the coefficients of a polynomial and its roots (in the case of a cubic: α, β, γ) can be solved like simultaneous equations. I started trying to solve them like simultaneous equations but it almost immediately became too much for me to handle.

$f(x)=ax^3+bx^2+cx+d$

$f(α)=0$

$f(β)=0$

$f(γ)=0$

α+β+γ=−$\frac{b}{a}$

αβ+βγ+γα=$\frac{c}{a}$

αβγ=−$\frac{d}{a}$

Has anyone solved for each root in a cubic (and/or quartic) equation using these equations? If so, where can I find those solutions?

Answer 1

This is what I get when I tell Mathematica to solve the system:

Input:

Reduce[{x + y + z == -b/a, x y + y z + x z == c/a, x y z == -d/a}, {x,y,z}]

Output:

a != 0 && (x == Root[d + c #1 + b #1^2 + a #1^3 &, 1] || 
x == Root[d + c #1 + b #1^2 + a #1^3 &, 2] || 
x == Root[d + c #1 + b #1^2 + a #1^3 &, 3]) && (y == (-b - a x - 
 Sqrt[b^2 - 4 a c - 2 a b x - 3 a^2 x^2])/(2 a) || 
y == (-b - a x + Sqrt[b^2 - 4 a c - 2 a b x - 3 a^2 x^2])/(2 a)) &&
z == (-b - a x - a y)/a

Translation of the output: $a \neq 0$ and $x$ is a root of $a x^3 + bx^2+cx +d$, and

($y=\frac{-\sqrt{-3 a^2 x^2-2 a b x-4 a c+b^2}-a x-b}{2 a}$ or $y=\frac{\sqrt{-3 a^2 x^2-2 a b x-4 a c+b^2}-a x-b}{2 a}$) and $z=\frac{-a x-a y-b}{a}$.

Note that I didn't feed to the computer the information that this system of equations came from a cubic equation.

(For Mathematica users: I'm not using Solve because Solve will reduce it to a cubic and then use the hard-coded formula for the solutions of a cubic to express the roots).

So solving the system obtained from Vieta's formulas is equivalent to solving the cubic itself. In hindsight, this should have been expected, because the information that we started with is the information that characterizes $x$, $y$ and $z$ as the solutions of a cubic, and then we used this information to try and solve the cubic, whose solutions are precisely $x$, $y$ and $z$. But this is circular.

Compare with the quadratic case: in $ax^2+bx+c$ we have $x_1+x_2=-b/a$ and $x_1x_2=c/a$. To solve this system we express $x_2$ as $x_2=-b/a-x_1$ and substitute into the remaining equation. What we obtain is $-x_1^2-\frac ba x_1 -\frac ca$. But this is a quadratic again, equivalent to the one we started with.