Completing the square - Wikipedia

Simple attempts to combine the x2 and the bx rectangles into a larger square result in a missing corner. The term (b/2)2 added to each side of the above ... Completingthesquare FromWikipedia,thefreeencyclopedia Jumptonavigation Jumptosearch Methodforsolvingquadraticequations PlaymediaAnimationdepictingtheprocessofcompletingthesquare.(Details,animatedGIFversion) Inelementaryalgebra,completingthesquareisatechniqueforconvertingaquadraticpolynomialoftheform a x 2 + b x + c {\displaystyleax^{2}+bx+c} totheform a ( x − h ) 2 + k {\displaystylea(x-h)^{2}+k} forsomevaluesofhandk. Completingthesquareisusedin solvingquadraticequations, derivingthequadraticformula, graphingquadraticfunctions, evaluatingintegralsincalculus,suchasGaussianintegralswithalineartermintheexponent, findingLaplacetransforms. Inmathematics,completingthesquareisoftenappliedinanycomputationinvolvingquadraticpolynomials. Contents 1Overview 1.1Background 1.2Basicexample 1.3Generaldescription 1.4Non-moniccase 1.5Formula 1.5.1Scalarcase 1.5.2Matrixcase 2Relationtothegraph 3Solvingquadraticequations 3.1Irrationalandcomplexroots 3.2Non-moniccase 4Otherapplications 4.1Integration 4.2Complexnumbers 4.3Idempotentmatrix 5Geometricperspective 6Avariationonthetechnique 6.1Example:thesumofapositivenumberanditsreciprocal 6.2Example:factoringasimplequarticpolynomial 7References 8Externallinks Overview[edit] Background[edit] Theformulainelementaryalgebraforcomputingthesquareofabinomialis: ( x + p ) 2 = x 2 + 2 p x + p 2 . {\displaystyle(x+p)^{2}\,=\,x^{2}+2px+p^{2}.} Forexample: ( x + 3 ) 2 = x 2 + 6 x + 9 ( p = 3 ) ( x − 5 ) 2 = x 2 − 10 x + 25 ( p = − 5 ) . {\displaystyle{\begin{alignedat}{2}(x+3)^{2}\,&=\,x^{2}+6x+9&&(p=3)\\[3pt](x-5)^{2}\,&=\,x^{2}-10x+25\qquad&&(p=-5).\end{alignedat}}} Inanyperfectsquare,thecoefficientofxistwicethenumberp,andtheconstanttermisequaltop2. Basicexample[edit] Considerthefollowingquadraticpolynomial: x 2 + 10 x + 28. {\displaystylex^{2}+10x+28.} Thisquadraticisnotaperfectsquare,since28isnotthesquareof5: ( x + 5 ) 2 = x 2 + 10 x + 25. {\displaystyle(x+5)^{2}\,=\,x^{2}+10x+25.} However,itispossibletowritetheoriginalquadraticasthesumofthissquareandaconstant: x 2 + 10 x + 28 = ( x + 5 ) 2 + 3. {\displaystylex^{2}+10x+28\,=\,(x+5)^{2}+3.} Thisiscalledcompletingthesquare. Generaldescription[edit] Givenanymonicquadratic x 2 + b x + c , {\displaystylex^{2}+bx+c,} itispossibletoformasquarethathasthesamefirsttwoterms: ( x + 1 2 b ) 2 = x 2 + b x + 1 4 b 2 . {\displaystyle\left(x+{\tfrac{1}{2}}b\right)^{2}\,=\,x^{2}+bx+{\tfrac{1}{4}}b^{2}.} Thissquarediffersfromtheoriginalquadraticonlyinthevalueoftheconstant term.Therefore,wecanwrite x 2 + b x + c = ( x + 1 2 b ) 2 + k , {\displaystylex^{2}+bx+c\,=\,\left(x+{\tfrac{1}{2}}b\right)^{2}+k,} where k = c − b 2 4 {\displaystylek\,=\,c-{\frac{b^{2}}{4}}} .Thisoperationisknownascompletingthesquare. Forexample: x 2 + 6 x + 11 = ( x + 3 ) 2 + 2 x 2 + 14 x + 30 = ( x + 7 ) 2 − 19 x 2 − 2 x + 7 = ( x − 1 ) 2 + 6. {\displaystyle{\begin{alignedat}{1}x^{2}+6x+11\,&=\,(x+3)^{2}+2\\[3pt]x^{2}+14x+30\,&=\,(x+7)^{2}-19\\[3pt]x^{2}-2x+7\,&=\,(x-1)^{2}+6.\end{alignedat}}} Non-moniccase[edit] Givenaquadraticpolynomialoftheform a x 2 + b x + c {\displaystyleax^{2}+bx+c} itispossibletofactoroutthecoefficienta,andthencompletethesquarefortheresultingmonicpolynomial. Example: 3 x 2 + 12 x + 27 = 3 [ x 2 + 4 x + 9 ] = 3 [ ( x + 2 ) 2 + 5 ] = 3 ( x + 2 ) 2 + 3 ( 5 ) = 3 ( x + 2 ) 2 + 15 {\displaystyle{\begin{aligned}3x^{2}+12x+27&=3[x^{2}+4x+9]\\&{}=3\left[(x+2)^{2}+5\right]\\&{}=3(x+2)^{2}+3(5)\\&{}=3(x+2)^{2}+15\end{aligned}}} Thisprocessoffactoringoutthecoefficientacanfurtherbesimplifiedbyonlyfactorisingitoutofthefirst2terms.Theintegerattheendofthepolynomialdoesnothavetobeincluded. Example: 3 x 2 + 12 x + 27 = 3 [ x 2 + 4 x ] + 27 = 3 [ ( x + 2 ) 2 − 4 ] + 27 = 3 ( x + 2 ) 2 + 3 ( − 4 ) + 27 = 3 ( x + 2 ) 2 − 12 + 27 = 3 ( x + 2 ) 2 + 15 {\displaystyle{\begin{aligned}3x^{2}+12x+27&=3[x^{2}+4x]+27\\&{}=3\left[(x+2)^{2}-4\right]+27\\&{}=3(x+2)^{2}+3(-4)+27\\&{}=3(x+2)^{2}-12+27\\&{}=3(x+2)^{2}+15\end{aligned}}} Thisallowsthewritingofanyquadraticpolynomialintheform a ( x − h ) 2 + k . {\displaystylea(x-h)^{2}+k.} Formula[edit] Scalarcase[edit] Theresultofcompletingthesquaremaybewrittenasaformula.Inthegeneralcase,onehas[1] a x 2 + b x + c = a ( x − h ) 2 + k , {\displaystyleax^{2}+bx+c=a(x-h)^{2}+k,} with h = − b 2 a and k = c − a h 2 = c − b 2 4 a . {\displaystyleh=-{\frac{b}{2a}}\quad{\text{and}}\quadk=c-ah^{2}=c-{\frac{b^{2}}{4a}}.} Inparticular,whena=1,onehas x 2 + b x + c = ( x − h ) 2 + k , {\displaystylex^{2}+bx+c=(x-h)^{2}+k,} with h = − b 2 and k = c − h 2 = c − b 2 4 a . {\displaystyleh=-{\frac{b}{2}}\quad{\text{and}}\quadk=c-h^{2}=c-{\frac{b^{2}}{4a}}.} Bysolvingtheequation a ( x − h ) 2 + k = 0 {\displaystylea(x-h)^{2}+k=0} intermsof x − h , {\displaystylex-h,} andreorganizingtheresultingexpression,onegetsthequadraticformulafortherootsofthequadraticequation: x = − b ± b 2 − 4 a c 2 a . {\displaystylex={\frac{-b\pm{\sqrt{b^{2}-4ac}}}{2a}}.} Matrixcase[edit] Thematrixcaselooksverysimilar: x T A x + x T b + c = ( x − h ) T A ( x − h ) + k where h = − 1 2 A − 1 b and k = c − 1 4 b T A − 1 b {\displaystylex^{\mathrm{T}}Ax+x^{\mathrm{T}}b+c=(x-h)^{\mathrm{T}}A(x-h)+k\quad{\text{where}}\quadh=-{\frac{1}{2}}A^{-1}b\quad{\text{and}}\quadk=c-{\frac{1}{4}}b^{\mathrm{T}}A^{-1}b} where A {\displaystyleA} hastobesymmetric. If A {\displaystyleA} isnotsymmetrictheformulaefor h {\displaystyleh} and k {\displaystylek} have tobegeneralizedto: h = − ( A + A T ) − 1 b and k = c − h T A h = c − b T ( A + A T ) − 1 A ( A + A T ) − 1 b {\displaystyleh=-(A+A^{\mathrm{T}})^{-1}b\quad{\text{and}}\quadk=c-h^{\mathrm{T}}Ah=c-b^{\mathrm{T}}(A+A^{\mathrm{T}})^{-1}A(A+A^{\mathrm{T}})^{-1}b} . Relationtothegraph[edit] Graphsofquadraticfunctionsshiftedtotherightbyh=0,5,10,and15.Graphsofquadraticfunctionsshiftedupwardbyk=0,5,10,and15.Graphsofquadraticfunctionsshiftedupwardandtotherightby0,5,10,and15. Inanalyticgeometry,thegraphofanyquadraticfunctionisaparabolainthexy-plane.Givenaquadraticpolynomialoftheform a ( x − h ) 2 + k {\displaystylea(x-h)^{2}+k} thenumbershandkmaybeinterpretedastheCartesiancoordinatesofthevertex(orstationarypoint)oftheparabola.Thatis,histhex-coordinateoftheaxisofsymmetry(i.e.theaxisofsymmetryhasequationx=h),andkistheminimumvalue(ormaximumvalue,ifa  0andb > 0,maybeexpressedintermsofthesquareoftheabsolutevalueofacomplexnumber.Define z = a x + i b y . {\displaystylez={\sqrt{a}}\,x+i{\sqrt{b}}\,y.} Then | z | 2 = z z ∗ = ( a x + i b y ) ( a x − i b y ) = a x 2 − i a b x y + i b a y x − i 2 b y 2 = a x 2 + b y 2 , {\displaystyle{\begin{aligned}|z|^{2}&{}=zz^{*}\\&{}=({\sqrt{a}}\,x+i{\sqrt{b}}\,y)({\sqrt{a}}\,x-i{\sqrt{b}}\,y)\\&{}=ax^{2}-i{\sqrt{ab}}\,xy+i{\sqrt{ba}}\,yx-i^{2}by^{2}\\&{}=ax^{2}+by^{2},\end{aligned}}} so a x 2 + b y 2 + c = | z | 2 + c . {\displaystyleax^{2}+by^{2}+c=|z|^{2}+c.} Idempotentmatrix[edit] AmatrixMisidempotentwhenM2=M.Idempotentmatricesgeneralizetheidempotentpropertiesof0and1.Thecompletionofthesquaremethodofaddressingtheequation a 2 + b 2 = a , {\displaystylea^{2}+b^{2}=a,} showsthatsomeidempotent2×2matricesareparametrizedbyacircleinthe(a,b)-plane: Thematrix ( a b b 1 − a ) {\displaystyle{\begin{pmatrix}a&b\\b&1-a\end{pmatrix}}} willbeidempotentprovided a 2 + b 2 = a , {\displaystylea^{2}+b^{2}=a,} which,uponcompletingthesquare,becomes ( a − 1 2 ) 2 + b 2 = 1 4 . {\displaystyle(a-{\tfrac{1}{2}})^{2}+b^{2}={\tfrac{1}{4}}.} Inthe(a,b)-plane,thisistheequationofacirclewithcenter(1/2,0)andradius1/2. Geometricperspective[edit] Considercompletingthesquarefortheequation x 2 + b x = a . {\displaystylex^{2}+bx=a.} Sincex2representstheareaofasquarewithsideoflengthx,andbxrepresentstheareaofarectanglewithsidesbandx,theprocessofcompletingthesquarecanbeviewedasvisualmanipulationofrectangles. Simpleattemptstocombinethex2andthebxrectanglesintoalargersquareresultinamissingcorner.Theterm(b/2)2addedtoeachsideoftheaboveequationispreciselytheareaofthemissingcorner,whencederivestheterminology"completingthesquare". Avariationonthetechnique[edit] Asconventionallytaught,completingthesquareconsistsofaddingthethirdterm,v 2to u 2 + 2 u v {\displaystyleu^{2}+2uv} togetasquare.Therearealsocasesinwhichonecanaddthemiddleterm,either2uvor−2uv,to u 2 + v 2 {\displaystyleu^{2}+v^{2}} togetasquare. Example:thesumofapositivenumberanditsreciprocal[edit] Bywriting x + 1 x = ( x − 2 + 1 x ) + 2 = ( x − 1 x ) 2 + 2 {\displaystyle{\begin{aligned}x+{1\overx}&{}=\left(x-2+{1\overx}\right)+2\\&{}=\left({\sqrt{x}}-{1\over{\sqrt{x}}}\right)^{2}+2\end{aligned}}} weshowthatthesumofapositivenumberxanditsreciprocalisalwaysgreaterthanorequalto2.Thesquareofarealexpressionisalwaysgreaterthanorequaltozero,whichgivesthestatedbound;andhereweachieve2justwhenxis1,causingthesquaretovanish. Example:factoringasimplequarticpolynomial[edit] Considertheproblemoffactoringthepolynomial x 4 + 324. {\displaystylex^{4}+324.} Thisis ( x 2 ) 2 + ( 18 ) 2 , {\displaystyle(x^{2})^{2}+(18)^{2},} sothemiddletermis2(x2)(18) = 36x2.Thusweget x 4 + 324 = ( x 4 + 36 x 2 + 324 ) − 36 x 2 = ( x 2 + 18 ) 2 − ( 6 x ) 2 = adifferenceoftwosquares = ( x 2 + 18 + 6 x ) ( x 2 + 18 − 6 x ) = ( x 2 + 6 x + 18 ) ( x 2 − 6 x + 18 ) {\displaystyle{\begin{aligned}x^{4}+324&{}=(x^{4}+36x^{2}+324)-36x^{2}\\&{}=(x^{2}+18)^{2}-(6x)^{2}={\text{adifferenceoftwosquares}}\\&{}=(x^{2}+18+6x)(x^{2}+18-6x)\\&{}=(x^{2}+6x+18)(x^{2}-6x+18)\end{aligned}}} (thelastlinebeingaddedmerelytofollowtheconventionofdecreasingdegreesofterms). Thesameargumentshowsthat x 4 + 4 a 4 {\displaystylex^{4}+4a^{4}} isalwaysfactorizableas x 4 + 4 a 4 = ( x 2 + 2 a x + 2 a 2 ) ( x 2 − 2 a x + 2 a 2 ) {\displaystylex^{4}+4a^{4}=(x^{2}+2ax+2a^{2})(x^{2}-2ax+2a^{2})} (AlsoknownasSophieGermain'sidentity). References[edit] ^Narasimhan,Revathi(2008).Precalculus:BuildingConceptsandConnections.CengageLearning.pp. 133–134.ISBN 0-618-41301-4.,SectionFormulafortheVertexofaQuadraticFunction,page133–134,figure2.4.8 Algebra1,Glencoe,ISBN 0-07-825083-8,pages539–544 Algebra2,Saxon,ISBN 0-939798-62-X,pages214–214,241–242,256–257,398–401 Externallinks[edit] WikimediaCommonshasmediarelatedtoCompletingthesquare. CompletingthesquareatPlanetMath. 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