Editing Quadratic formula (section)

== {{anchor|Derivations}} Other derivations ==
Any generic method or algorithm for solving quadratic equations can be applied to an equation with symbolic coefficients and used to derive some closed-form expression equivalent to the quadratic formula. Alternative methods are sometimes simpler than completing the square, and may offer interesting insight into other areas of mathematics.

=== Completing the square by Śrīdhara's method ===
Instead of dividing by {{tmath|a}} to isolate {{tmath|\textstyle x^2\!}}, it can be slightly simpler to multiply by {{tmath|4a}} instead to produce {{tmath|\textstyle (2ax)^2\!}}, which allows us to complete the square without need for fractions. Then the steps of the derivation are:<ref name="Hoehn1975">{{citation |last=Hoehn |first=Larry |title=A More Elegant Method of Deriving the Quadratic Formula |journal=The Mathematics Teacher |year=1975 |volume=68 |issue=5 |pages=442–443 |jstor=27960212 |doi=10.5951/MT.68.5.0442 }}</ref>

# Multiply each side by {{tmath|4a}}.
# Add {{tmath|\textstyle b^2 - 4ac}} to both sides to complete the square.
# Take the square root of both sides.
# Isolate {{tmath|x}}.

Applying this method to a generic quadratic equation with symbolic coefficients yields the quadratic formula:

<math display=block>\begin{align}
ax^2 + bx + c &= 0 \\[3mu]
4 a^2 x^2 + 4abx + 4ac &= 0 \\[3mu]
4 a^2 x^2 + 4abx + b^2 &= b^2 - 4ac \\[3mu]
(2ax + b)^2 &= b^2 - 4ac \\[3mu]
2ax + b &= \pm \sqrt{b^2 - 4ac} \\[5mu]
x &= \dfrac{-b\pm\sqrt{b^2 - 4ac }}{2a}. \vphantom\bigg)
\end{align}</math>

This method for completing the square is ancient and was known to the 8th–9th century Indian mathematician [[Śrīdhara]].<ref>Starting from a quadratic equation of the form {{tmath|1=\textstyle ax^2 + bx = c}}, Śrīdhara's method, as quoted by [[ Bhāskara II]] (c. 1150): "Multiply both sides of the equation by a number equal to four times the [coefficient of the] square, and add to them a number equal to the square of the original [coefficient of the] unknown quantity. [Then extract the root.]". {{pb}} {{harvnb|Smith|1923|loc=[https://archive.org/details/historyofmathema0001davi/page/446 {{pgs|446}}]}}</ref> Compared with the modern standard method for completing the square, this alternate method avoids fractions until the last step and hence does not require a rearrangement after step 3 to obtain a common denominator in the right side.<ref name=Hoehn1975/>

===By substitution===
Another derivation uses a [[change of variables]] to eliminate the linear term. Then the equation takes the form {{tmath|1=\textstyle u^2 = s}} in terms of a new variable {{tmath|u}} and some constant expression {{tmath|s}}, whose roots are then {{tmath|1= u = \pm \sqrt s}}.

By substituting {{tmath|1= x = u - \tfrac{b}{2a} }} into {{tmath|1=\textstyle ax^2 + bx + c = 0}}, expanding the products and combining like terms, and then solving for {{tmath|\textstyle u^2\!}}, we have:

<math display=block>\begin{align}
a\left(u-\frac{b}{2a}\right)^2 + b\left(u-\frac{b}{2a}\right) + c &=0 \\[5mu]
a\left(u^2-\frac{b}{a}u+\frac{b^2}{4a^2}\right) + b\left(u-\frac{b}{2a}\right) + c &= 0 \\[5mu]
au^2 - bu + \frac{b^2}{4a} + bu - \frac{b^2}{2a}+c &= 0 \\[5mu]
au^2 + \frac{4ac - b^2}{4a} &= 0 \\[5mu]
u^2 &= \frac{b^2 - 4ac}{4a^2}.
\end{align}</math>

Finally, after taking a square root of both sides and substituting the resulting expression for {{tmath|u}} back into {{tmath|1= x = u - \tfrac{b}{2a},}} the familiar quadratic formula emerges:

<math display=block>
x = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a}.
</math>

===By using algebraic identities===
The following method was used by many historical mathematicians:<ref>{{citation |doi = 10.1080/00207390802642237|title = The legacy of Leonhard Euler – a tricentennial tribute | year = 2009|last1 = Debnath|first1 = Lokenath|journal = International Journal of Mathematical Education in Science and Technology|volume = 40|issue = 3|pages = 353–388 | s2cid = 123048345}}</ref>

Let the roots of the quadratic equation {{tmath|1=\textstyle ax^2 + bx + c = 0}} be {{tmath|\alpha}} and {{tmath|\beta}}. The derivation starts from an identity for the square of a difference (valid for any two complex numbers), of which we can take the square root on both sides:

<math display=block>\begin{align}
(\alpha - \beta)^2 &= (\alpha + \beta)^2 - 4 \alpha\beta \\[3mu]
\alpha - \beta &= \pm\sqrt{(\alpha + \beta)^2 - 4 \alpha\beta} .
\end{align}</math>

Since the coefficient {{tmath|a \neq 0}}, we can divide the quadratic equation by {{tmath|a}} to obtain a [[monic polynomial|monic]] polynomial with the same roots. Namely,

<math display=block>
x^2 + \frac{b}{a}x + \frac{c}{a}
= (x - \alpha)(x - \beta)
= x^2 - (\alpha + \beta)x + \alpha\beta .
</math>

This implies that the sum {{tmath|1= \alpha + \beta = -\tfrac ba}} and the product {{tmath|1= \alpha\beta = \tfrac ca}}. Thus the identity can be rewritten:

<math display=block>
\alpha - \beta
= \pm\sqrt{\left(-\frac{b}{a}\right)^2-4\frac{c}{a}}
= \pm\frac{\sqrt{b^2 - 4ac}}{a} .
</math>

Therefore,

<math display=block>\begin{align}
\alpha &= \tfrac12(\alpha + \beta) + \tfrac12(\alpha - \beta)
        = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}, \\[10mu]
\beta  &= \tfrac12(\alpha + \beta) - \tfrac12(\alpha - \beta)
        = -\frac{b}{2a} \mp \frac{\sqrt{b^2 - 4ac}}{2a}.
\end{align}</math>

The two possibilities for each of {{tmath|\alpha}} and {{tmath|\beta}} are the same two roots in opposite order, so we can combine them into the standard quadratic equation:
<math display=block> x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} .</math>

===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.