Lihat sumber Rumus kuadrat

{{Periksa terjemahan}}

{{short description|solusi dari persamaan kuadrat}}
[[Berkas:Quadratic roots.svg|alt=Akar dari fungsi kuadrat|thumb|231x231px|Fungsi kuadrat dengan akar '' x '' = 1 dan '' x '' = 4.]]
Dalam [[aljabar elementer]], '''rumus kuadrat''' adalah rumus yang memberikan solusi untuk sebuah [[persamaan kuadrat]]. Ada cara lain untuk menyelesaikan persamaan kuadrat selain menggunakan rumus kuadrat, seperti [[faktorisasi]] (pemfaktoran langsung, pengelompokan, [[Faktorisasi#Metode AC|metode AC]]), [[menyelesaikan suatu kuadrat]], [[Grafik suatu fungsi|membuat atau menggambar grafik]],dan lain sebagainya.<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>

Diberikan persamaan kuadrat umum dalam bentuk

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

dengan <math>x</math> mewakili suatu variabel yang tidak diketahui. Variabel <math>a</math>, <math>b</math>, dan <math>c</math> mewakili [[konstanta]] dengan <math>a \ne 0</math>, rumus kuadratnya adalah:

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

dimana [[Tanda plus dan minus|tanda plus-minus "±"]] menunjukkan bahwa persamaan kuadrat memiliki dua solusi.<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> Dengan menulisnya secara terpisah, maka diperoleh:

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

Masing-masing dari dua solusi ini juga disebut [[Akar fungsi|akar]] dari persamaan kuadrat. Secara geometris, akar-akar tersebut mewakili nilai <math>x</math> di mana ''suatu '' [[parabola]] <math>y = ax^2 + bx + c</math>, memotong sumbu <math>x</math>.<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>

Selain menjadi rumus yang memberikan nilai nol dari suatu parabola, rumus kuadrat juga dapat digunakan untuk mengidentifikasi sumbu simetri 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> dan jumlah bilangan [[Bilangan real|real]] nol yang terdapat di persamaan kuadrat.<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>

== Perumusan yang setara ==
Rumus kuadrat juga dapat ditulis sebagai:

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

yang dapat disederhanakan menjadi:

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

Versi rumus ini cocok jika menggunakan akar kompleks, dalam hal ini ekspresi di luar akar kuadrat akan menjadi bagian riil, dan akar kuadrat ekspresi bagian imajiner. Ekspresi di dalam akar kuadrat adalah diskriminan.

=== Metode Muller ===
Rumus kuadrat yang kurang dikenal, yang digunakan di [[metode Muller]] dan yang dapat ditemukan dari [[rumus Vieta]], memberikan akar yang sama melalui persamaan:

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

=== Formulasi berdasarkan parameter alternatif ===
Parameterisasi standar dari persamaan kuadrat adalah 
:<math>ax^2+bx+c=0\ \ .</math>

Beberapa sumber, terutama yang lebih tua, menggunakan parameterisasi alternatif dari persamaan kuadrat seperti 
: <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>
atau 
: <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>

Parameter alternatif ini menghasilkan bentuk yang sedikit berbeda untuk solusi, tetapi yang sebaliknya setara dengan parametriisasi standar.

== {{anchor|Turunan lainnya}} Turunan rumus ==
Banyak metode berbeda untuk mendapatkan rumus kuadrat tersedia dalam literatur. Yang standar adalah aplikasi sederhana dari teknik [[menyelesaikan persegi]].<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): "Rumus kuadrat adalah metode paling umum untuk menyelesaikan persamaan kuadrat dan diturunkan dari metode umum lain: menyelesaikan kuadrat."</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> Metode alternatif terkadang lebih sederhana daripada menyelesaikan kuadrat, dan mungkin menawarkan wawasan menarik tentang bidang matematika lainnya.
<!--
===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 quadratic 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>''.

Completing the square can also be accomplished by a sometimes 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 derivation of the quadratic formula is ancient and was known in India 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 avoids fractions and squared 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/>

====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.{{citation needed|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>{{Cite journal |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}}</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 ("[[permutation|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 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.
-->

== Sejarah perkembangan ==
Metode paling awal untuk menyelesaikan persamaan kuadrat adalah geometri. Tablet paku Babilonia berisi soal yang dapat direduksi menjadi pemecahan persamaan kuadrat.<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>

Ahli matematika Yunani [[Euclid]] (sekitar 300 SM) menggunakan metode geometris untuk menyelesaikan persamaan kuadrat di Buku 2 dari '' [[Elemen Euklides|Elemen]] '', sebuah risalah matematika yang berpengaruh.<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://archive.org/details/historyofmathema0002smit|url-access=registration|year=1958|publisher=Courier Dover Publications|isbn=978-0-486-20430-7|page=[https://archive.org/details/historyofmathema0002smit/page/380 380]}}</ref> Dalam karyanya '' [[Arithmetica]] '', ahli matematika Yunani [[Diophantus]] (sekitar tahun 250 M) memecahkan persamaan kuadrat dengan metode aljabar yang lebih dikenali daripada aljabar geometris Euklides.<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://archive.org/details/historyofmathema0002smit|url-access=registration|year=1958|publisher=Courier Dover Publications|isbn=0-486-20429-4|page=[https://archive.org/details/historyofmathema0002smit/page/134 134]}}</ref>

Matematikawan India [[Brahmagupta]] (597–668 M) secara eksplisit mendeskripsikan rumus kuadrat dalam risalahnya '' [[Brāhmasphuṭasiddhānta]] '' yang diterbitkan pada 628 M,<ref name=Bradley>Bradley, Michael. ''The Birth of Mathematics: Ancient Times to 1300'', p. 86 (Infobase Publishing 2006).</ref> tetapi ditulis dengan kata-kata, bukan simbol.<ref>Mackenzie, Dana. ''The Universe in Zero Words: The Story of Mathematics as Told through Equations'', p. 61 (Princeton University Press, 2012).</ref> Solusi persamaan kuadratnya {{math|1=''ax''<sup>2</sup> + ''bx'' = ''c''}} adalah sebagai berikut: "Untuk bilangan absolut dikalikan dengan empat kali [koefisien] kuadrat, tambahkan kuadrat dari [koefisien] suku tengah; akar kuadratnya, dikurangi [koefisien] suku tengah, dibagi dua kali [koefisien] persegi adalah nilainya."<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>
Ini sama dengan:

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

Ahli matematika Persia abad ke-9 [[Muḥammad bin Mūsā al-Khwārizmī]] memecahkan persamaan kuadrat secara aljabar.<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> Rumus kuadrat yang mencakup semua kasus pertama kali diperoleh oleh [[Simon Stevin]] pada tahun 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> Pada tahun 1637 [[René Descartes]] menerbitkan '' [[La Géométrie]] '' berisi kasus khusus dari rumus kuadrat dalam bentuk yang kita kenal sekarang.<ref>{{Cite book|url=http://archive.org/details/TheGeometry|title=The Geometry|last=Rene Descartes|language=English}}</ref>

== Penggunaan yang signifikan ==

—Dalam proses --
<!--
=== Signifikansi geometris ===
{{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.
-->

== Lihat pula ==
* [[Diskriminan]]
* [[Teorema dasar aljabar]]
* [[Rumus Vieta]]
* [[Pemaktoran]]
* [[Persamaan Kuadrat]]
* [[Kuadratik]]

== Referensi ==
{{Reflist|30em}}

{{Polinomial}}

[[Kategori:Aljabar elementer]]
[[Kategori:Persamaan]]