View source for Quadratic formula

{{short description|solution of the quadratic equation}}
[[File:Quadratic roots.svg|alt=Roots of a quadratic function|thumb|231x231px|A quadratic function with roots ''x''&nbsp;=&nbsp;1 and ''x''&nbsp;=&nbsp;4.]]
In [[elementary algebra]], the '''quadratic formula''' is a formula which provides the solution(s) to a [[quadratic equation]]. There are [[Quadratic_equation#Solving the quadratic equation|other ways]] of solving a quadratic equation instead of using the quadratic formula, such as [[Factorization|factoring]] (direct factoring, grouping, AC method), [[completing the square]], [[Graph of a function|graphing]] and [[Quadratic equation#Alternative methods of root calculation|others]].<ref>{{Cite web|url=https://mathvault.ca/quadratic-factorisation/|title=Quadratic Factorisation: The Complete Guide|date=2016-03-13|website=Math Vault|language=en-US|access-date=2019-11-10}}</ref>

Given a general quadratic equation of the form

:<math>ax^2+bx+c=0</math>

with <math>x</math> representing an unknown, <math>a</math>, <math>b</math> and <math>c</math> representing [[Constant term|constant]]s with <math>a \ne 0</math>, the quadratic formula is: 

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

where the [[plus-minus sign|plus-minus symbol "±"]] indicates that the quadratic equation has two solutions.<ref>{{Citation|last=Sterling|first=Mary Jane|title=Algebra I For Dummies|year=2010|publisher=Wiley Publishing|isbn=978-0-470-55964-2|url=https://books.google.com/books?id=2toggaqJMzEC&pg=PA219&dq=quadratic+formula#v=onepage&q=quadratic%20formula&f=false|page=219}}</ref> Written separately, they become:

:<math> x_1=\frac{-b + \sqrt {b^2-4ac}}{2a}\quad\text{and}\quad x_2=\frac{-b - \sqrt {b^2-4ac}}{2a}</math>

Here, each of these two solutions is also called a [[Zero of a function|root]] of the quadratic equation (in fact, one can verify that each solution satisfies the quadratic equation by inserting them into the equation itself). Geometrically, these roots represent the <math>x</math> values at which ''any'' [[parabola]], explicitly given as <math>y = ax^2 + bx + c</math>, crosses the <math>x</math>-axis.<ref>{{Cite web|url=https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:quadratic-formula-a1/a/quadratic-formula-explained-article|title=Understanding the quadratic formula|website=Khan Academy|language=en|access-date=2019-11-10}}</ref>

As well as being a formula that yields the zeros of any parabola, the quadratic formula can also be used to identify the axis of symmetry of the parabola<ref>{{Cite web|url=https://www.mathwarehouse.com/geometry/parabola/axis-of-symmetry.php|title=Axis of Symmetry of a Parabola. How to find axis from equation or from a graph. To find the axis of symmetry ...|website=www.mathwarehouse.com|access-date=2019-11-10}}</ref>, and the number of [[Real number|real]] zeros the quadratic equation contains.<ref>{{Cite web|url=https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:quadratic-formula-a1/a/discriminant-review|title=Discriminant review|website=Khan Academy|language=en|access-date=2019-11-10}}</ref>

==Equivalent formulations==
The quadratic formula may also be written as:

:<math>x = \frac{-b}{2a} \pm \sqrt{\frac{b^2-4ac}{4a^2}} \ \ ,</math>

which may be simplified to:

:<math>x = -\frac{b}{2a} \pm \sqrt{\left(\frac{b}{2a}\right)^2-\frac{c}{a}} \ \ .</math>

This version of the formula is convenient when complex roots are involved, in which case the expression outside the square root will be the real part—and the square root expression the imaginary part. The expression inside the square root is a discriminant.

===Muller's method===
A lesser known quadratic formula, which is used in [[Muller's method]] and which can be found from [[Vieta's formulas]], provides the same roots via the equation:

:<math>x=\frac{-2c}{b\mp\sqrt{b^2-4ac}} = \frac{2c}{-b \pm \sqrt {b^2-4ac\ }}\ \ .</math>

===Formulations based on alternative parametrizations===
The standard parametrization of the quadratic equation is 
:<math>ax^2+bx+c=0\ \ .</math>

Some sources, particularly older ones, use alternative parameterizations of the quadratic equation such as 
: <math>ax^2 - 2b_1 x + c = 0</math>, where <math>b_1 = -b/2</math>,<ref name="kahan">{{Citation |first=Willian |last=Kahan |title=On the Cost of Floating-Point Computation Without Extra-Precise Arithmetic |url=http://www.cs.berkeley.edu/~wkahan/Qdrtcs.pdf |date=November 20, 2004 |accessdate=2012-12-25}}</ref>
or 
: <math>ax^2 + 2b_2 x + c = 0</math>, where <math>b_2 = b/2</math>.<ref>{{Citation |url=http://www.proofwiki.org/wiki/Quadratic_Formula |title=Quadratic Formula |journal=Proof Wiki |accessdate=2016-10-08}}</ref>

These alternative parametrizations result in slightly different forms for the solution, but which are otherwise equivalent to the standard parametrization.

==Derivations of the formula==
Many different methods to derive the quadratic formula are available in the literature. The standard one is a simple application of the [[completing the square]] technique.<ref>{{citation
|title=Schaum's Outline of Theory and Problems of Elementary Algebra
|first1=Barnett
|last1=Rich
|first2=Philip
|last2=Schmidt
|publisher=The McGraw–Hill Companies
|year=2004
|isbn=0-07-141083-X
|url=https://books.google.com/books?id=8PRU9cTKprsC}}, [https://books.google.com/books?id=8PRU9cTKprsC&pg=PA291 Chapter 13 §4.4, p. 291]</ref><ref>Li, Xuhui. ''An Investigation of Secondary School Algebra Teachers' Mathematical Knowledge for Teaching Algebraic Equation Solving'', p. 56 (ProQuest, 2007): "The quadratic formula is the most general method for solving quadratic equations and is derived from another general method: completing the square."</ref><ref>Rockswold, Gary. ''College algebra and trigonometry and precalculus'', p. 178 (Addison Wesley, 2002).</ref><ref>Beckenbach, Edwin et al. ''Modern college algebra and trigonometry'', p. 81 (Wadsworth Pub. Co., 1986).</ref> Alternative methods are sometimes simpler than completing the square, and may offer interesting insight into other areas of mathematics.

===By using the 'completing the square' technique===
====Standard method====
Divide the quadratic equation by <math>a</math>, which is allowed because <math>a</math> is non-zero:

:<math>x^2 + \frac{b}{a} x + \frac{c}{a}=0\ \ .</math>

Subtract {{math|{{sfrac|''c''|''a''}}}} from both sides of the equation, yielding:
:<math>x^2 + \frac{b}{a} x= -\frac{c}{a}\ \ .</math>

The quadratic equation is now in a form to which the method of [[completing the square]] is applicable. In fact, by adding a constant to both sides of the equation such that the left hand side becomes a complete square, the equation becomes:

:<math>x^2+\frac{b}{a}x+\left( \frac{b}{2a} \right)^2 =-\frac{c}{a}+\left( \frac{b}{2a} \right)^2\ \ ,</math>

which produces:

:<math>\left(x+\frac{b}{2a}\right)^2=-\frac{c}{a}+\frac{b^2}{4a^2}\ \ .</math>

Accordingly, after rearranging the terms on the right hand side to have a common denominator, we obtain:

:<math>\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\ \ .</math>

The square has thus been completed. Taking the [[square root]] of both sides yields the following equation:

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

In which case, isolating the <math>x</math> would give the quadratic formula:

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

There are many alternatives of this derivation with minor differences, mostly concerning the manipulation of <math>a</math>.

====Method 2====
The majority of algebra texts published over the last several decades teach [[completing the square]] using the sequence presented earlier: 

# Divide each side by <math>a</math> to make the polynomial [[Monic polynomial|monic]].
# Rearrange.
# Add <math>\textstyle \left( \frac{b}{2a} \right)^2</math> to both sides to complete the square.
# Rearrange the terms in the right side to have a common denominator.
# Take the square root of both sides.
# Isolate ''<math>x</math>''.

As pointed out by Larry Hoehn in 1975, completing the square can be accomplished by a shorter and simpler sequence:<ref name="Hoehn1975">{{cite journal |last=Hoehn |first=Larry |title=A More Elegant Method of Deriving the Quadratic Formula |journal=The Mathematics Teacher |year=1975 |volume=68 |issue=5 |page=442&ndash;443}}</ref>

# Multiply each side by ''<math>4a</math>'',
# Rearrange.
# Add ''<math>b^2</math>'' to both sides to complete the square.
# Take the square root of both sides.
# Isolate ''<math>x</math>''.

In which case, the quadratic formula can also be derived as follows:

:<math>\begin{align}
ax^2+bx+c &= 0 \\
4 a^2 x^2 + 4abx + 4ac &= 0 \\
4 a^2 x^2 + 4abx &= -4ac \\
4 a^2 x^2 + 4abx + b^2 &= b^2 - 4ac \\
(2ax + b)^2 &= b^2 - 4ac \\
2ax + b &= \pm \sqrt{b^2-4ac} \\
2ax &= -b \pm \sqrt{b^2-4ac} \\ 
x &= \frac{-b\pm\sqrt{b^2-4ac }}{2a}\ \ .
\end{align}</math>

This actually represents an ancient derivation of the quadratic formula and was known to the Hindus at least as far back as 1025.<ref name=Smith1958>{{cite book|last=Smith|first=David E.|title=History of Mathematics, Vol. II|year=1958|publisher=Dover Publications|isbn=0486204308|page=446}}</ref> Compared with the derivation in standard usage, this alternate derivation is simpler and shorter, as it avoids fractions and squared fractions until the last step and hence it does not require a rearrangement after step 3 to obtain a common denominator in the right side.<ref name=Hoehn1975/>

====Method 3====

Similar to Method 1, divide each side by ''<math>a</math>'' to make left-hand-side polynomial [[Monic polynomial|monic]] (i.e., the coefficient of ''<math>x^2</math>'' becomes ''1'').

:<math>x^2 + \frac{b}{a} x + \frac{c}{a}=0\ \ .</math>

Write the equation in a more compact and easier-to-handle format:

:<math>x^2 + Bx + C = 0</math>

where ''<math>\textstyle B = \frac{b}{a}</math>'' and ''<math>\textstyle C = \frac{c}{a}</math>''.

Complete the square by adding ''<math>\textstyle \frac{B^2}{4}</math>'' to the first two terms and subtracting it from the third term:

:<math>\left(x+\frac{B}{2}\right)^2 + \left( C - \frac{B^2}{4}\right) = 0</math>

Rearrange the left side into a [[difference of two squares]]: 

:<math>\left(x+\frac{B}{2}\right)^2 - \left( \sqrt{\frac{B^2}{4} - C}\right)^2 = 0</math>

and factor it:

:<math>\left(x+\frac{B}{2} + \sqrt{\frac{B^2}{4} - C}\right) \left(x+\frac{B}{2} - \sqrt{\frac{B^2}{4} - C}\right) = 0</math>

which implies that either

:<math>x+\frac{B}{2} + \sqrt{\frac{B^2}{4} - C} = 0</math>

or

:<math>x+\frac{B}{2} - \sqrt{\frac{B^2}{4} - C} = 0</math>

Each of these two equations is linear and can be solved for ''<math>x</math>'', obtaining:<blockquote><math>x = -\frac{B}{2} - \sqrt{\frac{B^2}{4} - C}</math>   or  <math>x = -\frac{B}{2} + \sqrt{\frac{B^2}{4} - C}</math></blockquote>By re-expressing ''<math>B</math>'' and ''<math>C</math>'' back into ''<math>\textstyle\frac{b}{a}</math>'' and ''<math>\textstyle\frac{c}{a}</math>'', respectively, the quadratic formula can then be obtained.{{cn|date=July 2019}}

===By substitution===
Another technique is solution by [[substitution (algebra)|substitution]].<ref>Joseph J. Rotman. (2010). Advanced modern algebra (Vol. 114). American Mathematical Soc. Section 1.1</ref> In this technique, we substitute ''<math>x = y+m</math>'' into the quadratic to get:

:<math>a(y+m)^2 + b(y+m) + c =0\ \ .</math>

Expanding the result and then collecting the powers of ''<math>y</math>'' produces:

:<math>ay^2 + y(2am + b) + \left(am^2+bm+c\right) = 0\ \ .</math>

We have not yet imposed a second condition on ''<math>y</math>'' and ''<math>m</math>'', so we now choose ''<math>m</math>'' so that the middle term vanishes. That is, ''<math>2am + b = 0</math>'' or ''<math>\textstyle m = \frac{-b}{2a}</math>''. Subtracting the constant term from both sides of the equation (to move it to the right hand side) and then dividing by ''<math>a</math>'' gives:

:<math>y^2=\frac{-\left(am^2+bm+c\right)}{a}\ \ .</math>

Substituting for ''<math>m</math>'' gives:

:<math>y^2=\frac{-\left(\frac{b^2}{4a}+\frac{-b^2}{2a}+c\right)}{a}=\frac{b^2-4ac}{4a^2}\ \ .</math>

Therefore,

:<math>y=\pm\frac{\sqrt{b^2-4ac}}{2a}</math>

By re-expressing ''<math>y</math>'' in terms of ''<math>x</math>'' using the formula ''<math>\textstyle x = y + m = y - \frac{b}{2a}</math>'' , the usual quadratic formula can then be obtained:

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

===By using algebraic identities===
The following method was used by many historical mathematicians:<ref>Debnath, L. (2009). [https://www.tandfonline.com/doi/abs/10.1080/00207390802642237 The legacy of Leonhard Euler–a tricentennial tribute]. International Journal of Mathematical Education in Science and Technology, 40(3), 353–388. Section 3.6</ref>

Let the roots of the standard quadratic equation be {{math|''r''<sub>1</sub>}} and {{math|''r''<sub>2</sub>}}. The derivation starts by recalling the identity:

:<math>(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1r_2\ \ .</math>

Taking the square root on both sides, we get:

:<math>r_1 - r_2 = \pm\sqrt{(r_1 + r_2)^2 - 4r_1r_2}\ \ .</math>

Since the coefficient {{math|1=''a'' ≠ 0}}, we can divide the standard equation by {{math|''a''}} to obtain a quadratic polynomial having the same roots. Namely,

:<math> x^2 + \frac{b}{a}x + \frac{c}{a} = (x - r_1)(x-r_2) = x^2 - (r_1 + r_2)x + r_1 r_2\ \ .</math>

From this we can see that the sum of the roots of the standard quadratic equation is given by {{math|−{{sfrac|''b''|''a''}}}}, and the product of those roots is given by {{math|{{sfrac|''c''|''a''}}}}.
<!--
We know that the sum of roots of the standard quadratic equation is given by {{math|−{{sfrac|''b''|''a''}}}}:

:<math>\frac{-b + \sqrt{b^2 - 4ac}}{2a} + \frac{-b - \sqrt{b^2 - 4ac}}{2a} = \frac{-b - b + \sqrt{b^2-4ac} - \sqrt{b^2-4ac}}{2a} = \frac{-2b}{2a} = -\frac{b}{a}</math>\ \ .

Additionally, the product is given by {{math|{{sfrac|''c''|''a''}}}}:

:<math>\frac{-b + \sqrt{b^2 - 4ac}}{2a} \cdot \frac{-b - \sqrt{b^2 - 4ac}}{2a} = \frac{(-b + \sqrt{b^2 - 4ac})(-b - \sqrt{b^2 - 4ac})}{4a^2} = \frac{b^2 - (b^2 - 4ac)}{4a^2} = \frac{4ac}{4a^2} = \frac{c}{a}\ \.</math>
-->
Hence the identity can be rewritten as:

:<math>r_1 - r_2 = \pm\sqrt{\left(-\frac{b}{a}\right)^2-4\frac{c}{a}} = \pm\sqrt{\frac{b^2}{a^2} - \frac{4ac}{a^2}} = \pm\frac{\sqrt{b^2-4ac}}{a}\ \ .</math>

Now, 

:<math>r_1 = \frac{(r_1 + r_2) + (r_1 - r_2)}{2} = \frac{-\frac{b}{a} \pm \frac{\sqrt{b^2 - 4ac}}{a}}{2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\ \ .</math>

Since {{math|''r''{{sub|2}} {{=}} −''r''{{sub|1}} − {{sfrac|''b''|''a''}}}}, if we take

:<math>r_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}</math>

then we obtain

:<math>r_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}\ \ ;</math>

and if we instead take

:<math>r_1 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}</math>

then we calculate that

:<math>r_2 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}\ \ .</math>

Combining these results by using the standard shorthand ±, we have that the solutions of the quadratic equation are given by:

:<math> 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
}}, [https://books.google.com/books?id=fcp9IiZd3tQC&pg=PA134#PPA134,M1 §6.2, p. 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'' more than on rearranging the original equation. Given a monic quadratic polynomial

:<math>x^2+px+q\ \ ,</math> 

assume that it factors as

:<math>x^2+px+q=(x-\alpha)(x-\beta)\ \ ,</math>

Expanding yields

:<math>x^2+px+q=x^2-(\alpha+\beta)x+\alpha \beta\ \ ,</math>

where {{math|''p'' {{=}} −(''α'' + ''β'')}} and {{math|''q'' {{=}} ''αβ''}}.

Since the order of multiplication does not matter, one can switch {{math|''α''}} and {{math|''β''}} and the values of {{math|''p''}} and {{math|''q''}} will not change: one can say that {{math|''p''}} and {{math|''q''}} are [[symmetric polynomials]] in {{math|''α''}} and {{math|''β''}}. In fact, they are the [[elementary symmetric polynomials]] – any symmetric polynomial in {{math|''α''}} and {{math|''β''}} can be expressed in terms of {{math|''α'' + ''β''}} and {{math|''αβ''}} The Galois theory approach to analyzing and solving polynomials is: given the coefficients of a polynomial, which are symmetric functions in the roots, can one "break the symmetry" and recover the roots? Thus solving a polynomial of degree {{math|''n''}} is related to the ways of rearranging ("permuting") {{math|''n''}} terms, which is called the [[symmetric group]] on {{math|''n''}} letters, and denoted {{math|''S{{sub|n}}''}}. For the quadratic polynomial, the only way to rearrange two terms is to swap them ("[[Transposition (mathematics)|transpose]]" them), and thus solving a quadratic polynomial is simple.

To find the roots {{math|''α''}} and {{math|''β''}}, consider their sum and difference:

:<math>\begin{align}
r_1 &= \alpha + \beta\\
r_2 &= \alpha - \beta\ \ .
\end{align}</math>

These are called the ''Lagrange resolvents'' of the polynomial; notice that one of these depends on the order of the roots, which is the key point. One can recover the roots from the resolvents by inverting the above equations:

:<math>\begin{align}
\alpha &= \textstyle{\frac{1}{2}}\left(r_1+r_2\right)\\
\beta  &= \textstyle{\frac{1}{2}}\left(r_1-r_2\right)\ \ .
\end{align}</math>

Thus, solving for the resolvents gives the original roots.

Now {{math|''r''{{sub|1}} {{=}} ''α'' + ''β''}} is a symmetric function in {{math|''α''}} and {{math|''β''}}, so it can be expressed in terms of {{math|''p''}} and {{math|''q''}}, and in fact {{math|''r''{{sub|1}} {{=}} −''p''}} as noted above. But {{math|''r''{{sub|2}} {{=}} ''α'' − ''β''}} is not symmetric, since switching {{math|''α''}} and {{math|''β''}} yields {{math|−''r''{{sub|2}} {{=}} ''β'' − ''α''}} (formally, this is termed a [[Group action (mathematics)|group action]] of the symmetric group of the roots). Since {{math|''r''{{sub|2}}}} is not symmetric, it cannot be expressed in terms of the coefficients {{math|''p''}} and {{math|''q''}}, as these are symmetric in the roots and thus so is any polynomial expression involving them. Changing the order of the roots only changes {{math|''r''{{sub|2}}}} by a factor of −1, and thus the square {{math|''r''{{sub|2}}{{sup|2}} {{=}} (''α'' − ''β''){{sup|2}}}} is symmetric in the roots, and thus expressible in terms of {{math|''p''}} and {{math|''q''}}. Using the equation

:<math>(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\ \ </math>

yields

:<math>r_2^2 = p^2 - 4q\ \ </math>

and thus

:<math>r_2 = \pm \sqrt{p^2 - 4q}\ \ </math>

If one takes the positive root, breaking symmetry, one obtains:

:<math>\begin{align}
r_1 &= -p\\
r_2 &= \sqrt{p^2 - 4q}
\end{align}</math>
and thus
:<math>\begin{align}
\alpha &= \tfrac12\left(-p+\sqrt{p^2 - 4q}\right)\\
\beta  &= \tfrac12\left(-p-\sqrt{p^2 - 4q}\right)\ \ .
\end{align}</math>
Thus the roots are
:<math>\textstyle{\frac{1}{2}}\left(-p \pm \sqrt{p^2 - 4q}\right)</math>
which is the quadratic formula. Substituting {{math|''p'' {{=}} {{sfrac|''b''|''a''}}, ''q'' {{=}} {{sfrac|''c''|''a''}}}} yields the usual form for when a quadratic is not monic. The resolvents can be recognized as {{math|{{sfrac|''r''{{sub|1}}|2}} {{=}} {{sfrac|−''p''|2}} {{=}} {{sfrac|−''b''|2''a''}}}} being the vertex, and {{math|''r''{{sub|2}}{{sup|2}} {{=}} ''p''{{sup|2}} − 4''q''}} is the discriminant (of a monic polynomial).

A similar but more complicated method works for [[cubic equation]]s, where one has three resolvents and a quadratic equation (the "resolving polynomial") relating {{math|''r''{{sub|2}}}} and {{math|''r''{{sub|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.<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.

===By extrema===
Knowing the value of {{math|''x''}} in the functional extreme point makes it possible to solve only for the increase (or decrease) needed in {{math|''x''}} to solve the quadratic equation. This method first uses differentiation to find the {{math|''x''}} value at the extremum, called {{math|''x''<sub>ext</sub>}}. We then solve for the value, {{math|''q''}}, that ensures that {{math|''f''(''x''<sub>ext</sub> + ''q'') {{=}} 0}}. While this may not be the most intuitive method, it ensures that the mathematics is straightforward.

:<math>\begin{align}
f(x) & =ax^2+bx+c \\[5pt]
\frac{\partial f}{\partial x} & =2ax+b\ \ .
\end{align}</math>

Setting the above differential to zero will give us the extrema of the quadratic function
:<math>x_\text{ext}=\frac{-b}{2a}\ \ .</math>

We define {{math|''q''}} as follows:

:<math> q=x_0-x_\text{ext}\ \ ,</math>

Here {{math|''x''<sub>0</sub>}} is the value of {{math|''x''}} that solves the quadratic equation. The sum of {{math|''x''<sub>ext</sub>}} and the variable of interest, {{math|''q''}}, is plugged in to the quadratic equation

:<math>\begin{align}
a\left(\frac{-b}{2a}+q\right)^2+b\left(\frac{-b}{2a}+q\right)+c&=0 \\[5pt]
\left(\frac{-b}{2a}+q\right)^2+\frac{b}{a}\left(\frac{-b}{2a} + q \right) +\frac{c}{a}&=0 &&(a\neq 0)\\[5pt]
\frac{b^2}{4a^2}+q^2-\frac{bq}{a}-\frac{b^2}{2a^2}+\frac {bq} a + \frac{c}{a}&=0 \\[5pt]
\frac{-b^2}{4a^2}+q^2+\frac{c}{a}&=0 \\[5pt]
q^2&=\frac{b^2-4ac}{4a^2} \\[5pt]
q&=\frac{\pm\sqrt{b^2-4ac}}{2a}\ \ .
\end{align}</math>

The value of {{math|''x''}} in the extreme point is then added to both sides of the equation
:<math> x_0=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\ \ 
.</math>

This gives the quadratic formula. This way one avoids the technique of completing the square and much more complicated mathematics is not needed. Note this solution is very similar to solving deriving the formula by substitution.

===By splitting into real and imaginary parts===
{{quadratic_function_graph_complex_roots.svg}}
Consider the equation
:<math> a z^2 + b z + c = 0, </math>
where <math>z = x + iy</math> is a complex number, and ''a'', ''b'', and ''c'' are real numbers. Then
:<math>\begin{align}
  a (x + iy)^2 + b (x + iy) + c &= 0 \\
  a (x^2 - y^2 + i2xy) + b(x + iy) + c &= 0. 
 \end{align}</math>

This splits into two equations, the real part:
:<math> a(x^2 - y^2) + bx + c = 0</math>
and the imaginary part:
:<math> 2axy + by = 0.</math>

Assuming that <math>y \ne 0,</math> divide the second equation by ''y'':
:<math> 2ax + b = 0</math>
and solve for ''x'':
:<math>x = \frac{-b}{2a}.</math>

Substitute this value for ''x'' into the first equation and solve for ''y'':
:<math>\begin{align}
   a \left( \frac{b^2}{4 a^2} - y^2 \right) - \frac{b^2}{2 a} + c &= 0\\
   \frac{b^2}{4a} - ay^2 - \frac{2b^2}{4a} + c &= 0\\
   \frac{-b^2}{4a} - ay^2 + c &= 0\\
   ay^2 + \frac{b^2}{4a} - c &= 0\\
   y^2 &= \frac{c}{a} - \frac{b^2}{4a^2}\\
   y &= \pm \sqrt{\frac{c}{a} - \frac{b^2}{4a^2}} = \frac{\pm \sqrt{ca - \frac{b^2}{4}}}{a} = \frac{\pm \sqrt{4ca - b^2}}{2a}.
 \end{align}</math>

Since <math>z = x + iy,</math>
:<math>\begin{align}
   z &= \frac{-b}{2a} \pm \frac{i\sqrt{4ac - b^2}}{2a}\\[5pt]
   z &= \frac{-b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
 \end{align}</math>
Even though ''y'' was assumed to be non-zero, this last formula holds for any roots of the original equation.

==Historical development==
The earliest methods for solving quadratic equations were geometric. Babylonian cuneiform tablets contain problems reducible to solving quadratic equations.<ref>{{cite book|last=Irving|first=Ron|title=Beyond the Quadratic Formula|url=https://books.google.com/books?id=CV_UInCRO38C&pg=PA39|year=2013|publisher=MAA|isbn=978-0-88385-783-0|page=34}}</ref> The Egyptian [[Berlin Papyrus 6619|Berlin Papyrus]], dating back to the [[Middle Kingdom of Egypt|Middle Kingdom]] (2050 BC to 1650 BC), contains the solution to a two-term quadratic equation.<ref>{{cite book|title=The Cambridge Ancient History Part 2 Early History of the Middle East|url=https://books.google.com/books?id=slR7SFScEnwC&pg=PA530|year=1971|publisher=Cambridge University Press|isbn=978-0-521-07791-0|page=530}}</ref>
[[File:Scuola di atene 07.jpg|thumb|250px|[[Euclid]] in [[Raphael|Raphael Sanzio]]'s ''[[The School of Athens|School of Athens]]'']]
The Greek mathematician [[Euclid]] (circa 300 BC) used geometric methods to solve quadratic equations in Book 2 of his ''[[Euclid's Elements|Elements]]'', an influential mathematical treatise.<ref name=quad/> Rules for quadratic equations appear in the Chinese ''[[The Nine Chapters on the Mathematical Art]]'' circa 200 BC.<ref name=Aitken>{{cite web|last=Aitken|first=Wayne|title=A Chinese Classic: The Nine Chapters|url=http://public.csusm.edu/aitken_html/m330/china/ninechapters.pdf|publisher=Mathematics Department, California State University|accessdate=28 April 2013}}</ref><ref>{{cite book|last=Smith|first=David Eugene|title=History of Mathematics|url=https://books.google.com/books?id=uTytJGnTf1kC&pg=PA380|year=1958|publisher=Courier Dover Publications|isbn=978-0-486-20430-7|page=380}}</ref> In his work ''[[Arithmetica]]'', the Greek mathematician [[Diophantus]] (circa 250 BC) solved quadratic equations with a method more recognizably algebraic than the geometric algebra of Euclid.<ref name=quad>{{cite book|last=Irving|first=Ron|title=Beyond the Quadratic Formula|url=https://books.google.com/books?id=CV_UInCRO38C&pg=PA39|year=2013|publisher=MAA|isbn=978-0-88385-783-0|page=39}}</ref> His solution gives only one root, even when both roots are positive.<ref>{{cite book|last=Smith|first=David Eugene|title=History of Mathematics|url=https://books.google.com/?id=12qdOZ0gsWoC&pg=PA134&dq#v=onepage&q=&f=false|year=1958|publisher=Courier Dover Publications|isbn=0-486-20429-4|page=134}}</ref>

The Indian mathematician [[Brahmagupta]] (597–668 AD) explicitly described the quadratic formula in his treatise ''[[Brāhmasphuṭasiddhānta]]'' published in 628 AD,<ref name=Bradley>Bradley, Michael. ''The Birth of Mathematics: Ancient Times to 1300'', p. 86 (Infobase Publishing 2006).</ref> but written in words instead of symbols.<ref>Mackenzie, Dana. ''The Universe in Zero Words: The Story of Mathematics as Told through Equations'', p. 61 (Princeton University Press, 2012).</ref> His solution of the quadratic equation {{math|1=''ax''<sup>2</sup> + ''bx'' = ''c''}} was as follows: "To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the value."<ref name=Stillwell2004>{{cite book |last=Stillwell |first=John |title=Mathematics and Its History (2nd ed.) |year=2004 |publisher=Springer |isbn=0-387-95336-1|page=87}}</ref>
This is equivalent to:

:<math>x = \frac{\sqrt{4ac+b^2}-b}{2a}\ \ .</math>

The 9th-century Persian mathematician [[Muḥammad ibn Mūsā al-Khwārizmī]] solved quadratic equations algebraically.<ref>{{cite book|last=Irving|first=Ron|title=Beyond the Quadratic Formula|url=https://books.google.com/books?id=CV_UInCRO38C&pg=PA39|year=2013|publisher=MAA|isbn=978-0-88385-783-0|page=42}}</ref> The quadratic formula covering all cases was first obtained by [[Simon Stevin]] in 1594.<ref>{{Citation |title=The Principal Works of Simon Stevin, Mathematics |volume=II-B |first1=D. J. |last1=Struik |first2=Simon |last2=Stevin |publisher=C. V. Swets & Zeitlinger |year=1958 |page=470 |url=http://www.dwc.knaw.nl/pub/bronnen/Simon_Stevin-%5bII_B%5d_The_Principal_Works_of_Simon_Stevin,_Mathematics.pdf}}</ref> In 1637 [[René Descartes]] published ''[[La Géométrie]]'' containing special cases of the quadratic formula in the form we know today.{{fact|date=January 2017}} The first appearance of the general solution in the modern mathematical literature appeared in an 1896 paper by [[Henry Heaton]].<ref name="heaton-1896">Heaton, H. (1896) ''[https://www.jstor.org/stable/info/2971099 A Method of Solving Quadratic Equations]'', [[American Mathematical Monthly]] '''3'''(10), 236–237.</ref>

==Significant uses==
===Geometric significance===
{{quadratic_equation_graph_key_points.svg}}
In terms of coordinate geometry, a parabola is a curve whose {{math|(''x'', ''y'')}}-coordinates are described by a second-degree polynomial, i.e. any equation of the form:

:<math>y =p(x) = a_2x^2 + a_1x +a_0\ \ ,</math>

where {{math|''p''}} represents the polynomial of degree 2 and {{math|''a''<sub>0</sub>, ''a''<sub>1</sub>,}} and {{math|1=''a''<sub>2</sub> ≠ 0}} are constant coefficients whose subscripts correspond to their respective term's degree. The geometrical interpretation of the quadratic formula is that it defines the points on the {{math|''x''}}-axis where the parabola will cross the axis. Additionally, if the quadratic formula was looked at as two terms,

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

the [[axis of symmetry]] appears as the line {{math|1=''x'' = −{{sfrac|''b''|2''a''}}}}. The other term, {{math|{{sfrac|{{sqrt|''b''{{sup|2}} − 4''ac''}}|2''a''}}}}, gives the distance the zeros are away from the axis of symmetry, where the plus sign represents the distance to the right, and the minus sign represents the distance to the left. 

If this distance term were to decrease to zero, the value of the axis of symmetry would be the {{math|''x''}} value of the only zero, that is, there is only one possible solution to the quadratic equation. Algebraically, this means that {{math|{{sqrt|''b''{{sup|2}} − 4''ac''}} {{=}} 0}}, or simply {{math|''b''{{sup|2}} − 4''ac'' {{=}} 0}} (where the left-hand side is referred to as the ''discriminant''). This is one of three cases, where the discriminant indicates how many zeros the parabola will have. If the discriminant is positive, the distance would be non-zero, and there will be two solutions. However, there is also the case where the discriminant is less than zero, and this indicates the distance will be ''imaginary''{{snd}} or some multiple of the complex unit {{math|''i''}}, where {{math|''i'' {{=}} {{sqrt|−1}}}}{{snd}} and the parabola's zeros will be [[complex number]]s. The complex roots will be [[complex conjugate]]s, where the real part of the complex roots will be the value of the axis of symmetry. There will be no real values of {{math|''x''}} where the parabola crosses the {{math|''x''}}-axis.

===Dimensional analysis===
If the constants {{math|''a''}}, {{math|''b''}}, and/or {{math|''c''}} are not [[unitless]], then the units of {{math|''x''}} must be equal to the units of {{math|{{sfrac|''b''|''a''}}}}, due to the requirement that {{math|''ax''{{sup|2}}}} and {{math|''bx''}} agree on their units. Furthermore, by the same logic, the units of {{math|''c''}} must be equal to the units of {{math|{{sfrac|''b''{{sup|2}}|''a''}}}}, which can be verified without solving for {{math|''x''}}. This can be a powerful tool for verifying that a quadratic expression of [[physical quantities]] has been set up correctly, prior to solving this.

==See also==
* [[Discriminant]]
* [[Fundamental theorem of algebra]]
* [[Cubic Formula]]

==References==
{{Reflist|30em}}

{{Polynomials}}

[[Category:Elementary algebra]]
[[Category:Equations]]