stylistic:style_transfer, linguistic:paraphrasing
By Lagrange resolvents

An alternative way of deriving the quadratic formula is via the method of Lagrange resolvents, which is an early part of Galois theory. This method can be generalized to give the roots of cubic polynomials and quartic polynomials, and leads to Galois theory, which allows one to understand the solution of algebraic equations of any degree in terms of the symmetry group of their roots, the Galois group.

This approach focuses on the roots themselves rather than algebraically rearranging the original equation. Given a monic quadratic polynomial x 2 + p x + q {\displaystyle \textstyle x^{2}+px+q} assume that α {\displaystyle \alpha } and β {\displaystyle \beta } are the two roots. So the polynomial factors as

x 2 + p x + q = (x − α) (x − β) = x 2 − (α + β) x + α β {\displaystyle {\begin{aligned}x^{2}+px+q&=(x-\alpha)(x-\beta)\\&=x^{2}-(\alpha +\beta)x+\alpha \beta \end{aligned}}}

which implies p = − (α + β) {\displaystyle p=-(\alpha +\beta)} and q = α β {\displaystyle q=\alpha \beta }.

Since multiplication and addition are both commutative, exchanging the roots α {\displaystyle \alpha } and β {\displaystyle \beta } will not change the coefficients p {\displaystyle p} and q {\displaystyle q}: one can say that p {\displaystyle p} and q {\displaystyle q} are symmetric polynomials in α {\displaystyle \alpha } and β {\displaystyle \beta }. Specifically, they are the elementary symmetric polynomials – any symmetric polynomial in α {\displaystyle \alpha } and β {\displaystyle \beta } can be expressed in terms of α + β {\displaystyle \alpha +\beta } and α β {\displaystyle \alpha \beta } instead.

The Galois theory approach to analyzing and solving polynomials is to ask whether, given coefficients of a polynomial each of which is a symmetric function in the roots, one can "break" the symmetry and thereby recover the roots. Using this approach, solving a polynomial of degree n {\displaystyle n} is related to the ways of rearranging (" permuting ") n {\displaystyle n} terms, called the symmetric group on n {\displaystyle n} letters and denoted S n {\displaystyle S_{n}}. For the quadratic polynomial, the only ways to rearrange two roots are to either leave them be or to transpose them, so solving a quadratic polynomial is simple.

To find the roots α {\displaystyle \alpha } and β {\displaystyle \beta }, consider their sum and difference:

r 1 = α + β, r 2 = α − β. {\displaystyle r_{1}=\alpha +\beta,\quad r_{2}=\alpha -\beta.}

These are called the Lagrange resolvents of the polynomial, from which the roots can be recovered as

α = 1 2 (r 1 + r 2), β = 1 2 (r 1 − r 2). {\displaystyle \alpha ={\tfrac {1}{2}}(r_{1}+r_{2}),\quad \beta ={\tfrac {1}{2}}(r_{1}-r_{2}).}

Because r 1 = α + β {\displaystyle r_{1}=\alpha +\beta } is a symmetric function in α {\displaystyle \alpha } and β {\displaystyle \beta }, it can be expressed in terms of p {\displaystyle p} and q, {\displaystyle q,} specifically r 1 = − p {\displaystyle r_{1}=-p} as described above. However, r 2 = α − β {\displaystyle r_{2}=\alpha -\beta } is not symmetric, since exchanging α {\displaystyle \alpha } and β {\displaystyle \beta } yields the additive inverse − r 2 = β − α {\displaystyle -r_{2}=\beta -\alpha }. So r 2 {\displaystyle r_{2}} cannot be expressed in terms of the symmetric polynomials. However, its square r 2 2 = (α − β) 2 {\displaystyle \textstyle r_{2}^{2}=(\alpha -\beta)^{2}} is symmetric in the roots, expressible in terms of p {\displaystyle p} and q {\displaystyle q}. Specifically r 2 2 = (α − β) 2 = {\displaystyle \textstyle r_{2}^{2}=(\alpha -\beta)^{2}={}} (α + β) 2 − 4 α β = {\displaystyle \textstyle (\alpha +\beta)^{2}-4\alpha \beta ={}} p 2 − 4 q {\displaystyle \textstyle p^{2}-4q}, which implies r 2 = ± p 2 − 4 q {\displaystyle \textstyle r_{2}=\pm {\sqrt {p^{2}-4q}}}. Taking the positive root "breaks" the symmetry, resulting in

r 1 = − p, r 2 = p 2 − 4 q {\displaystyle r_{1}=-p,\qquad r_{2}={\textstyle {\sqrt {p^{2}-4q}}}}

from which the roots α {\displaystyle \alpha } and β {\displaystyle \beta } are recovered as

x = 1 2 (r 1 ± r 2) = 1 2 (− p ± p 2 − 4 q) {\displaystyle x={\tfrac {1}{2}}(r_{1}\pm r_{2})={\tfrac {1}{2}}{\bigl (}{-p}\pm {\textstyle {\sqrt {p^{2}-4q}}}\,{\bigr)}}

which is the quadratic formula for a monic polynomial.

Substituting p = b / a {\displaystyle p=b/a}, q = c / a {\displaystyle q=c/a} yields the usual expression for an arbitrary quadratic polynomial. The resolvents can be recognized as

1 2 r 1 = − 1 2 p = − b 2 a, r 2 2 = p 2 − 4 q = b 2 − 4 a c a 2, {\displaystyle {\tfrac {1}{2}}r_{1}=-{\tfrac {1}{2}}p=-{\frac {b}{2a}},\qquad r_{2}^{2}=p_{2}-4q={\frac {b^{2}-4ac}{a^{2}}},}

respectively the vertex and the discriminant of the monic polynomial.

A similar but more complicated method works for cubic equations, which have three resolvents and a quadratic equation (the "resolving polynomial") relating r 2 {\displaystyle r_{2}} and r 3 {\displaystyle r_{3}}, which one can solve by the quadratic equation, and similarly for a quartic equation (degree 4), whose resolving polynomial is a cubic, which can in turn be solved. The same method for a quintic equation yields a polynomial of degree 24, which does not simplify the problem, and, in fact, solutions to quintic equations in general cannot be expressed using only roots.