Editing Quadratic formula (section)

===By Lagrange resolvents===
{{details|Lagrange resolvents}}
An alternative way of deriving the quadratic formula is via the method of [[Lagrange resolvents]],<ref name=Clark>Clark, A. (1984). ''Elements of abstract algebra''. Courier Corporation. p. 146.</ref> which is an early part of [[Galois theory]].<ref name="efei">{{citation |title=Elliptic functions and elliptic integrals |first1=Viktor |last1=Prasolov |first2=Yuri |last2=Solovyev |publisher=AMS Bookstore |year=1997 |isbn=978-0-8218-0587-9 |url=https://books.google.com/books?id=fcp9IiZd3tQC&pg=PA134 |page=134 }}</ref>
This method can be generalized to give the roots of [[cubic polynomial]]s and [[quartic polynomial]]s, 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 {{tmath|\textstyle x^2 + px + q}} assume that {{tmath|\alpha}} and {{tmath|\beta}} are the two roots. So the polynomial factors as

<math display=block>\begin{align}
x^2+px+q &= (x-\alpha)(x-\beta) \\[3mu]
         &= x^2-(\alpha+\beta)x+\alpha\beta
\end{align}</math>

which implies {{tmath|1= p = -(\alpha + \beta)}} and {{tmath|1= q = \alpha\beta}}.

Since multiplication and addition are both [[commutative property|commutative]], exchanging the roots {{tmath|\alpha}} and {{tmath|\beta}} will not change the coefficients {{tmath|p}} and {{tmath|q}}: one can say that {{tmath|p}} and {{tmath|q}} are [[symmetric polynomials]] in {{tmath|\alpha}} and {{tmath|\beta}}. Specifically, they are the [[elementary symmetric polynomials]] – any symmetric polynomial in {{tmath|\alpha}} and {{tmath|\beta}} can be expressed in terms of {{tmath|\alpha + \beta}} and {{tmath|\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 {{tmath|n}} is related to the ways of rearranging ("[[permutation|permuting]]") {{tmath|n}} terms, called the [[symmetric group]] on {{tmath|n}} letters and denoted {{tmath|S_n}}. For the quadratic polynomial, the only ways to rearrange two roots are to either leave them be or to [[Transposition (mathematics)|transpose]] them, so solving a quadratic polynomial is simple.

To find the roots {{tmath|\alpha}} and {{tmath|\beta}}, consider their sum and difference:

<math display=block>
r_1 = \alpha + \beta, \quad r_2 = \alpha - \beta .
</math>

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

<math display=block>
\alpha = \tfrac12 (r_1 + r_2), \quad \beta = \tfrac12(r_1 - r_2).
</math>

Because {{tmath|1= r_1 = \alpha + \beta}} is a symmetric function in {{tmath|\alpha}} and {{tmath|\beta}}, it can be expressed in terms of {{tmath|p}} and {{tmath|q,}} specifically {{tmath|1= r_1 = -p}} as described above. However, {{tmath|1= r_2 = \alpha - \beta}} is not symmetric, since exchanging {{tmath|\alpha}} and {{tmath|\beta}} yields the additive inverse {{tmath|1= -r_2 = \beta - \alpha}}. So {{tmath|r_2}} cannot be expressed in terms of the symmetric polynomials. However, its square {{tmath|1=\textstyle r_2^2 = (\alpha - \beta)^2}} ''is'' symmetric in the roots, expressible in terms of {{tmath|p}} and {{tmath|q}}. Specifically {{tmath|1=\textstyle r_2^2 = (\alpha - \beta)^2 = {} }}{{wbr}}{{tmath|1=\textstyle (\alpha + \beta)^2 - 4\alpha\beta = {} }}{{wbr}}{{tmath|1=\textstyle p^2 - 4q}}, which implies {{tmath|1=\textstyle r_2 = \pm \sqrt{p^2 - 4q} }}. Taking the positive root "breaks" the symmetry, resulting in

<math display=block>
r_1 = -p, \qquad r_2 = {\textstyle \sqrt{p^2 - 4q}}
</math>

from which the roots {{tmath|\alpha}} and {{tmath|\beta}} are recovered as

<math display=block>
x = \tfrac12(r_1 \pm r_2)
= \tfrac{1}{2} \bigl({-p} \pm {\textstyle \sqrt{p^2 - 4q}}\,\bigr)
</math>

which is the quadratic formula for a monic polynomial.

Substituting {{tmath|1= p = b/a}}, {{tmath|1= q = c/a}} yields the usual expression for an arbitrary quadratic polynomial. The resolvents can be recognized as

<math display=block>
\tfrac12 r_1 = -\tfrac12p = -\frac{b}{2a}, \qquad
r_2^2 = p_2 - 4q = \frac{b^2 - 4ac}{a^2},
</math>

respectively the vertex and the discriminant of the monic polynomial.

A similar but more complicated method works for [[cubic equation]]s, which have three resolvents and a quadratic equation (the "resolving polynomial") relating {{tmath|r_2}} and {{tmath|r_3}}, which one can solve by the quadratic equation, and similarly for a [[quartic equation]] ([[degree of a polynomial|degree]] 4), whose resolving polynomial is a cubic, which can in turn be solved.<ref name=Clark/> 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.