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| [[File:Quadratic formula.svg|thumbnail|The quadratic formula]]
| | Hi! <br>My name is Kathrin and I'm a 22 years old girl from Germany.<br><br>Review my webpage ... [http://z1Ont.83Rpm.com/photo/26673 FIFA coin generator] |
| In [[elementary algebra|basic algebra]], the '''quadratic formula''' is the solution of the [[quadratic equation]]. There are other ways to solve the quadratic equation instead of using the quadratic formula, such as [[Factorization|factoring]], [[completing the square]], or [[Graph of a function|graphing]]. However, using the quadratic formula is often the most convenient way.
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| The quadratic equation is
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| :<math>ax^2+bx+c=0.</math>
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| Here ''x'' represents an unknown, and ''a'', ''b'', and ''c'' are [[Constant term|constant]]s with ''a'' not equal to 0. One can easily verify that the quadratic formula satisfies the quadratic equation, by inserting the former into the latter. Each of the solutions given by the quadratic formula is called a [[Zero of a function|root]] of the quadratic equation.
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| ==Derivation of the formula==
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| Once a student understands how to [[completing the square|complete the square]], then he or she can derive the quadratic formula.<ref>{{citation
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| |title=Schaum's Outline of Theory and Problems of Elementary Algebra
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| |first1=Barnett
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| |last1=Rich
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| |first2=Philip
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| |last2=Schmidt
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| |publisher=The McGraw–Hill Companies
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| |year=2004
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| |isbn=0-07-141083-X
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| |url=http://books.google.com/?id=8PRU9cTKprsC}}, [http://books.google.be/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> For that reason, the derivation is sometimes left as an exercise for the student, who can thereby experience rediscovery of this important formula.<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> The explicit derivation is as follows.
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| Divide the quadratic equation by ''a'', which is allowed because ''a'' is non-zero:
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| :<math>x^2 + \frac{b}{a} x + \frac{c}{a}=0.</math>
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| Subtract ''c''/''a'' from both sides of the equation, transforming it into the form
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| :<math>x^2 + \frac{b}{a} x= -\frac{c}{a}.</math>
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| [[File:Completing the Square for the Quadratic Eq.png|thumb|300px|One way to think of this derivation is in geometric terms. Lengths are in green, and area is in red.]]The quadratic equation is now in a form to which the method of completing the square can be applied. To "complete the square", add a constant to both sides of the equation such that the left hand side becomes a complete square:
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| :<math>x^2+\frac{b}{a}x+\left( \frac{1}{2}\frac{b}{a} \right)^2 =-\frac{c}{a}+\left( \frac{1}{2}\frac{b}{a} \right)^2,</math>
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| which produces
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| :<math>\left(x+\frac{b}{2a}\right)^2=-\frac{c}{a}+\frac{b^2}{4a^2}</math>
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| or (after rearranging the terms on the right hand side to have a common denominator)
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| :<math>\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}.</math>
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| The square has thus been completed, as shown in the figure. Taking the [[square root]] of both sides yields
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| :<math>x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac\ }}{2a}.</math>
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| Isolating ''x'' gives the quadratic formula:
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| :<math>x=\frac{-b\pm\sqrt{b^2-4ac\ }}{2a}.</math>
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| The [[plus-minus sign|plus-minus symbol "±"]] indicates that both
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| :<math> x=\frac{-b + \sqrt {b^2-4ac}}{2a}\quad\text{and}\quad x=\frac{-b - \sqrt {b^2-4ac}}{2a}</math>
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| are solutions of the quadratic equation.<ref>{{Citation|last=Sterling|first=Mary Jane|title=Algebra I For Dummies|year=2010|publisher=Wiley Publishing|isbn=978-0-470-55964-2|url=http://books.google.com/?id=2toggaqJMzEC&pg=PA219&dq=quadratic+formula#v=onepage&q=quadratic%20formula&f=false|page=219}}</ref> There are many alternatives of this derivation with minor differences, mostly concerning the manipulation of <math>a</math>.
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| Some sources, particularly older ones, use alternative parameterizations of the quadratic equation such as <math>ax^2-2bx+c=0</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+2bx+c=0</math>,<ref>{{Citation |url=http://www.proofwiki.org/wiki/Quadratic_Equation#Also_defined_as |title=Quadratic Equation |journal=Proof Wiki |accessdate=2012-12-25}}</ref> where ''b'' has a magnitude one half of the more common one. These result in slightly different forms for the solution, but are otherwise equivalent.
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| ==Historical development==
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| [[File:Brahmagupta.jpg|thumb|upright|[[Brahmagupta]] described the quadratic formula more clearly than anyone had done before.]] The [[Indian mathematics|Indian mathematician]] [[Brahmagupta]] (597–668 AD) was first to clearly describe the quadratic formula, although prior civilizations had investigated quadratic equations, understood them fairly well, and developed methods for solving them.<ref name=Bradley>Bradley, Michael. ''The Birth of Mathematics: Ancient Times to 1300'', p. 86 (Infobase Publishing 2006).</ref> Brahmagupta's formula was written in words, instead of symbols, in his treatise ''[[Brāhmasphuṭasiddhānta]]'' published in 628 AD.<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>ax^2+bx=c</math> was as follows:
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| <blockquote>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. (''Brahmasphutasiddhanta'', Colebrook translation, 1817, page 346.)<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></blockquote>
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| This is equivalent to:
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| :<math>x = \frac{\sqrt{4ac+b^2}-b}{2a}.</math>
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| Mathematics Professor Elizabeth Stapel has explained what motivated the discovery of this formula:<ref>[http://www.purplemath.com/resume.htm Stapel, Elizabeth]. "[http://www.purplemath.com/modules/solvquad4.htm Solving Quadratic Equations: Solving with the Quadratic Formula]." Purplemath. Accessed 13 October 2013.</ref>
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| <blockquote>Somebody (possibly in seventh-century India) was solving a lot of quadratic equations by completing the square. At some point, he noticed that he was always doing the exact same steps in the exact same order for every equation. Taking advantage of one of the great powers and benefits of algebra (namely, the ability to deal with abstractions, rather than having to muck about with the numbers every single time), he made a formula out of what he'd been doing....</blockquote>
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| 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 the quadratic formula in the form we know today. 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) ''[http://www.jstor.org/stable/info/2971099 A Method of Solving Quadratic Equations]'', [[American Mathematical Monthly]] '''3'''(10), 236–237.</ref>
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| ==Importance of this solution==
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| Among the many equations that one encounters while studying [[algebra]], the quadratic formula is one of the most important, and is considered the most useful method of solving quadratic equations.<ref>Jahr, Cathy. ''Barron's How to Prepare for the Tennessee Gateway High School Exit Exam in Algebra'', p. 137 (Barron's Educational Series, 2005): "The Quadratic Formula is one of the most important formulas in mathematics because it is a method for solving all quadratic equations."</ref><ref>Heywood, Arthur. ''Intermediate algebra: lecture-lab'', p. 235 (Dickenson Pub. Co., 1975): "The quadratic formula is one of the most important formulas in mathematics, and we will now spend some time studying many different ways of using it."</ref> Unlike some other solution methods such as factoring, the quadratic formula can be used to solve any quadratic equation.<ref>Blanton, Floyd. ''Modern College Algebra'', p. 162 (McGraw–Hill, 1967): "The quadratic formula is the most powerful method for solving quadratics since it can be used to solve any quadratic."</ref><ref name=Smith>Smith, R. and Peterson, J. ''Introductory Technical Mathematics'', pp. 408–409 (Cengage Learning 2006): "The factoring method has limited application. Only certain quadratic equations can be solved by factoring. Completing the square…can be a rather long and complicated procedure and is seldom used in practical applications. [The] quadratic formula…is the most useful method for solving complete quadratic equations."</ref> Many equations that are not naively quadratic can be put into quadratic form, and solved using the quadratic formula.<ref>Banks, John. ''Elements of Algebra'', p. 97 (Allyn and Bacon, 1962): "The quadratic formula is one of the most useful formulas in elementary mathematics. You should be certain you know what it is and how to use it. Many other equations can be solved by first reducing them to quadratic form."</ref> For these reasons, it is often memorized.<ref>Larson, R. and Hodgkins A. ''College Algebra with Applications for Business and Life Sciences'', p. 104 (Cengage Learning 2009): "The Quadratic Formula is one of the most important formulas in algebra, and you should memorize it."</ref><ref>McConnell, John. ''Algebra'', p. 603 (Scott Foresman 1993): "The Quadratic Formula is one of the most famous formulas in all of mathematics. You should memorize it today."</ref>
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|
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| Completing the square also allows for the solution of all quadratics, as it is mathematically equivalent to the quadratic formula, but the quadratic formula gives a result without the need for so much algebraic manipulation. As such it is generally considered more practical to use the formula.<ref name=Smith /><ref>Payne, M. ''Intermediate Algebra'', p. 289 (West Publishing 1985): "While the method of completing the square may be used to solve quadratic equations, it is more involved than the quadratic formula, and is seldom used in practical work."</ref><ref>Davis, L. ''Technical Mathematics'', p. 174. (Merrill Publishing 1990): "You can use the quadratic formula, as well as completing the square, to solve any quadratic equation. However, you will find that the quadratic formula is easier to use."</ref><ref>Dugopolski, Mark. ''Algebra for College Students'', p. 541 (McGraw Hill 2006): "Any quadratic equation can be solved by completing the square or using the quadratic formula. Because the quadratic formula is usually faster, it is used more often than completing the square."</ref> However, completing the square is very useful for other purposes, such as putting the equations for [[conic section]]s into standard form.<ref name=sterling>Sterling, Mary. ''[http://books.google.com/books?id=-8VfOqh1CMEC&pg=PA60 CliffsStudySolver: Algebra II]'', p. 60 (Houghton Mifflin Harcourt 2012).</ref>
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| ==Other derivations==
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| A number of alternative derivations of the quadratic formula can be found in the literature which either (a) are simpler than the standard completing the square method, (b) represent interesting applications of other frequently used techniques in algebra, or (c) offer insight into other areas of mathematics.
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| ===Alternate method of completing the square===
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| The great majority of algebra texts published over the last several decades teach [[Quadratic equation#Completing the square|completing the square using the sequence presented earlier]]: (1) divide each side by ''a'', (2) rearrange, (3) then add the square of one-half of ''b/a''.
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| However, as pointed out by Larry Hoehn in 1975, completing the square can accomplished by a different sequence that leads to a simpler sequence of intermediate terms: (1) multiply each side by ''4a'', (2) rearrange, (3) then add <math>b^2</math>.<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=442–443}}</ref>
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| In other words, the quadratic formula can be derived as follows:
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| :<math>ax^2+bx+c=0</math>
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| :<math>4 a^2 x^2 + 4abx + 4ac=0</math>
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| :<math>4 a^2 x^2 + 4abx = -4ac</math>
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| :<math>4 a^2 x^2 + 4abx + b^2 = b^2 - 4ac </math>
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| :<math> (2ax + b)^2 = b^2 - 4ac </math>
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| :<math> 2ax + b = \pm \sqrt{b^2-4ac} </math>
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| :<math> 2ax = -b \pm \sqrt{b^2-4ac} </math>
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| :<math>x=\frac{-b\pm\sqrt{b^2-4ac\ }}{2a}</math>
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| This actually represents an ancient derivation of the quadratic formula, and was known to the Hindus at least as far back as 1025 A.D.<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 shorter, involves fewer computations with literal coefficients, avoids fractions until the last step, has simpler expressions, and uses simpler math. As Hoehn states, "it is easier 'to add the square of ''b''' than it is 'to add the square of half the coefficient of the ''x'' term'".<ref name=Hoehn1975/>
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| ===By substitution===
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| With this technique, we substitute <math>x=y+m</math> into the quadratic to get:
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| :<math>a(y+m)^2 + b(y+m) + c =0</math>
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| Expanding the result and then collecting the powers of <math>y</math> produces:
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| :<math>ay^2 + y(2am + b) + (am^2+bm+c) = 0</math>
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| We have not yet imposed a second condition on <math>y</math> and <math>m</math>, so we now choose ''m'' so that the middle term vanishes. That is, <math>2am+b=0</math> or <math>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 ''a'' gives:
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| :<math>y^2=\frac{-(am^2+bm+c)}{a}</math>
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| Substituting for <math>m</math> gives:
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| :<math>y^2=\frac{-(\frac{b^2}{4a}+\frac{-b^2}{2a}+c)}{a}=\frac{b^2-4ac}{4a^2}</math>
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| Therefore <math>y=\pm\frac{\sqrt{b^2-4ac}}{2a}</math>; substituting <math>x=y+m=y-\frac{b}{2a}</math> provides the quadratic formula.
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| ===By using algebraic identities===
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| {{confusing section|reason=identities used to derive the quadratic formula are derived from the quadratic formula|date=November 2013}}
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| Let the roots of the standard quadratic equation be <math>r_1</math> and <math>r_2</math>. At this point, we recall the identity:
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| :<math>r_1 - r_2 = \pm\sqrt{(r_1 + r_2)^2 - 4r_1r_2}</math>
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| We know that the sum of roots of the standard quadratic equation is given by <math>-\frac{b}{a}</math>:
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| :<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>
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| Additionally, the product is given by <math>\frac{c}{a}</math>:
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| :<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>
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| Hence the identity can be rewritten as:
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| :<math>r_1 - r_2 = \pm\sqrt{(-\frac{b}{a})^2-4\frac{c}{a}} = \pm\sqrt{\frac{b^2}{a^2} - \frac{4ac}{a^2}} = \pm\frac{\sqrt{b^2-4ac}}{a}</math>
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| Now, <math>x = \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>
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| ===By Lagrange resolvents===
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| {{details|Lagrange resolvents}}
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| An alternative way of deriving the quadratic formula is via the method of [[Lagrange resolvents]], which is an early part of [[Galois theory]].<ref name="efei">{{citation
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| |title=Elliptic functions and elliptic integrals
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| |first1=Viktor
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| |last1=Prasolov
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| |first2=Yuri
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| |last2=Solovyev
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| |publisher=AMS Bookstore
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| |year=1997
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| |isbn=978-0-8218-0587-9
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| |url=http://books.google.com/?id=fcp9IiZd3tQC
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| }}, [http://books.google.com/books?id=fcp9IiZd3tQC&pg=PA134#PPA134,M1 §6.2, p. 134]</ref>
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| 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]].
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| This approach focuses on the ''roots'' more than on rearranging the original equation.
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| Given a monic quadratic polynomial
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| :<math>x^2+px+q,</math>
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| assume that it factors as
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| :<math>x^2+px+q=(x-\alpha)(x-\beta),</math>
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| Expanding yields
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| :<math>x^2+px+q=x^2-(\alpha+\beta)x+\alpha \beta,</math>
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| where <math>p=-(\alpha+\beta)</math> and <math>q=\alpha \beta</math>.
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| Since the order of multiplication does not matter, one can switch <math>\alpha</math> and <math>\beta</math> and the values of ''p'' and ''q'' will not change: one says that ''p'' and ''q'' are [[symmetric polynomials]] in <math>\alpha</math> and <math>\beta</math>. In fact, they are the [[elementary symmetric polynomials]] – any symmetric polynomial in <math>\alpha</math> and <math>\beta</math> can be expressed in terms of <math>\alpha+\beta</math> and <math>\alpha\beta.</math> 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 ''n'' is related to the ways of rearranging ("permuting") ''n'' terms, which is called the [[symmetric group]] on ''n'' letters, and denoted <math>S_n.</math> 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.
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| To find the roots <math>\alpha</math> and <math>\beta,</math> consider their sum and difference:
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| :<math>\begin{align}
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| r_1 &= \alpha + \beta\\
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| r_2 &= \alpha - \beta.
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| \end{align}</math>
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| These are called the '''[[Lagrange resolvents]]''' of the polynomial;
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| notice that one of these depends on the order of the roots, which is the key point.
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| One can recover the roots from the resolvents by inverting the above equations:
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| :<math>\begin{align}
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| \alpha &= \textstyle{\frac{1}{2}}\left(r_1+r_2\right)\\
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| \beta &= \textstyle{\frac{1}{2}}\left(r_1-r_2\right).
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| \end{align}</math>
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| Thus, solving for the resolvents gives the original roots.
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| Formally, the resolvents are called the [[discrete Fourier transform]] (DFT) of order 2, and the transform can be expressed by the matrix <math>\left(\begin{smallmatrix}1 & 1\\ 1 & -1\end{smallmatrix}\right),</math> with inverse matrix <math>\left(\begin{smallmatrix}1/2 & 1/2\\ 1/2 & -1/2\end{smallmatrix}\right).</math> The transform matrix is also called the [[DFT matrix]] or [[Vandermonde matrix]].
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| Now <math>r_1=\alpha + \beta</math> is a symmetric function in <math>\alpha</math> and <math>\beta,</math> so it can be expressed in terms of ''p'' and ''q,'' and in fact <math>r_1 = -p,</math> as noted above. But <math>r_2=\alpha - \beta</math> is not symmetric, since switching <math>\alpha</math> and <math>\beta</math> yields <math>-r_2=\beta - \alpha</math> (formally, this is termed a [[group action]] of the symmetric group of the roots). Since <math>r_2</math> is not symmetric, it cannot be expressed in terms of the polynomials ''p'' and ''q'', as these are symmetric in the roots and thus so is any polynomial expression involving them. However, changing the order of the roots only changes <math>r_2</math> by a factor of <math>-1,</math> and thus the square <math>\scriptstyle r_2^2 = (\alpha - \beta)^2</math> ''is'' symmetric in the roots, and thus expressible in terms of ''p'' and ''q.'' Using the equation
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| :<math>(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\!</math>
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| yields
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| :<math>r_2^2 = p^2 - 4q\!</math>
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| and thus
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| :<math>r_2 = \pm \sqrt{p^2 - 4q}.\!</math>
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| If one takes the positive root, breaking symmetry, one obtains:
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| :<math>\begin{align}
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| r_1 &= -p\\
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| r_2 &= \sqrt{p^2 - 4q}
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| \end{align}</math>
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| and thus
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| :<math>\begin{align}
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| \alpha &= \textstyle{\frac{1}{2}}\left(-p+\sqrt{p^2 - 4q}\right)\\
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| \beta &= \textstyle{\frac{1}{2}}\left(-p-\sqrt{p^2 - 4q}\right)
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| \end{align}</math>
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| Thus the roots are
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| :<math>\textstyle{\frac{1}{2}}\left(-p \pm \sqrt{p^2 - 4q}\right)</math>
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| which is the quadratic formula. Substituting <math>\scriptstyle p=\tfrac{b}{a}, q=\tfrac{c}{a}\!</math> yields the usual form for when a quadratic is not monic. The resolvents can be recognized as <math>\scriptstyle \frac{r_1}{2} = \frac{-p}{2}=\frac{-b}{2a}\!</math> being the vertex, and <math>\scriptstyle r_2^2=p^2-4q\!</math> is the discriminant (of a monic polynomial).
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| A similar but more complicated method works for cubic equations, where one has three resolvents and a quadratic equation (the "resolving polynomial") relating <math>r_2</math> and <math>r_3,</math> which one can solve by the quadratic equation, and similarly for a quartic (degree 4) equation, whose resolving polynomial is a cubic, which can in turn be solved. However, 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.
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| ==See also==
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| *[[Discriminant]]
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| *[[Fundamental theorem of algebra]]
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| ==External links==
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| *[http://quadratic-formula-calculator.com/ Quadratic formula calculator]
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| *[http://mathworld.wolfram.com/QuadraticFormula.html Alternative formula (Wolfram)]
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| *[http://www.youtube.com/watch?v=2lbABbfU6Zc Quadratic formula put to music ("Pop Goes the Weasel")]
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| ==References==
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| {{reflist|30em}}
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| {{Polynomials}}
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| [[Category:Elementary algebra]]
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| [[Category:Equations]]
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