Many students enter ninth grade already familiar

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1 Delving eeper Henri Piiotto A New Path to the Quarati Formula Man stuents enter ninth grae alrea familiar with the quarati formula. Man others learn it in ninth grae. Some an even sing it! Unfortunatel the formula has little meaning for most stuents. For man the traitional erivation of the formula b ompleting the square (fig. ) if it is shown to them is more baffling than illuminating. As a teaher I value stuent unerstaning an earl on in m areer as an algebra teaher I foun this state of affairs isturbing. M first response was to have stuents omplete the square repeatel using numbers at first an then the parameters in the hope that this proess woul lea to unerstaning. Alas over time I realize that for man if not most of m stuents aitional smbol manipulation i not throw aitional light on the subjet. I neee to ome at this lak of unerstaning some other wa. Thus was launhe an on-an-off quest that le to multiple approahes to this subjet an a eeper unerstaning on m part. (Several stuent ativit sheets about this an losel relate material are available for ownloaing at org/math-e/parabolas/.) This epartment fouses on mathematis ontent that appeals to seonar shool teahers. It provies a forum that allows lassroom teahers to share the mathematis from their work with stuents their lassroom investigations an projets an This their epartment other eperienes. fouses on mathematis We enourage ontent submissions that appeals that pose to seonar an solve a shool novel or teahers. interesting It provies mathematis a forum problem that allows epan lassroom on onnetions teahers to among share their ifferent mathematis from their topis work present with stuents a general their metho lassroom for esribing investigations a mathematial an projets notion an mathematial or their solving other a eperienes. lass of problems We enourage elaborate submissions new insights that pose into an familiar solve seonar a novel or shool interesting mathematis mathematis or leave problem the reaer epan with on a onnetions mathematial among iea to ifferent epan. mathematial topis Delving present Deeper a general b metho aessing for mt.msubmit.net. esribing a mathematial notion Sen submissions or Delving solving a Deeper lass of an problems aept elaborate manusripts on in new ASCII insights or Wor into formats familiar onl. seonar shool mathematis or leave the reaer with a mathematial iea to epan. Sen Eite submissions b Al Cuoo to Delving auoo@e.org Deeper b aessing mt.msubmit.net. Center for Mathematis Euation Euation Development Center Newton Eite b MA Al 0458 Cuoo auoo@e.org Center for Mathematis Euation Newton MA 0458 E. Paul Golenberg pgolenberg@e.org Center E. Paul for Golenberg Mathematis pgolenberg@e.org Euation Euation Development Center Newton Center for MA Mathematis 0458 Euation Newton MA 0458 Beause I ame to high shool mathematis teahing after ten ears or so in elementar euation m quest initiall le through manipulatives (Piiotto 995 pp ). (See fig..) This approah is limite to the ase where a = an b > 0 but it generalizes reail to the traitional erivation an an provie a soli founation for unerstaning that erivation. As graphing tehnolog beame available the use of funtions as a wa to think about solving equations gaine urren. In this view the solution to a + b + = 0 is the set of -interepts of If 0 = a + b + omplete the square: 0 = + b a + a a = + b a a + b a = + b a + b a a + b = + b a + b = + b a b = + b a b = + b a ± b = + b a a = b a ± b. a So = b ± b. a Fig. Derivation of the quarati formula b ompleting the square Vol. 0 No. 6 Februar 008 Mathematis Teaher 473 Copright 008 The National Counil of Teahers of Mathematis In. All rights reserve. This material ma not be opie or istribute eletroniall or in an other format without written permission from NCTM.

2 + 8 9 = = 9 v = = 5 (h v) Fig. 3 Graph of a quarati with two real roots ( + 4) = = + 5 = or = 9 Fig. Manipulatives an be somewhat helpful to stuents tring to unerstan how to omplete the square v the funtion = a + b +. I was never onvine that all mathematis an be taught b wa of funtions but I beame intrigue b the possibilit of using that approah to erive the quarati formula an in fat I i fin suh a erivation (Wah an Piiotto 994 pp ). Here it is: If there are roots p an q the funtion an be written in fatore form: = a( p)( q) = a a(p + q) + apq () It follows that the prout of the roots is /a sine = apq an the sum of the roots is /a sine b = a(p + q). Let us use this information to fin (h v) the oorinates of the verte. The average of the roots h is /a. To fin v we substitute this epression into the formula simplif an get b a v = + 4. There are real roots if a > 0 an v 0 (a smiling parabola with verte on or below the -ais) or if a < 0 an v 0 (a frowning parabola with verte on or above the -ais). In both ases there are real roots if b is nonnegative; we have reisovere the isriminant. Fig. 4 The quarati from figure 3 shifte so that its verte is at the origin Let us onsier the ase where there are real roots an a > 0. (The ase a < 0 works similarl.) The graph looks something like that in figure 3. The -interepts are on either sie of the verte at a istane. So we have b = ± a. To fin shift the parabola so that its verte is at the origin (see fig. 4). Its equation is now simpl = a. The points that were the -interepts now have the oorinates (± v). It follows that Thus = b v = a or = a b b b an = ±. a a. 474 Mathematis Teaher Vol. 0 No. 6 Februar 008

3 In (a) the two interepts of the parabola are an 8 as an be seen in the fatore form of the equation. In (b) a retangle with area is shown in the stanar representation b using algebra manipulatives in this ase Lab Gear (Piiotto 995). The imensions of the retangle are ( + ) an ( + 8). The piees are one blok ten bloks an siteen bloks. Beause the area of the ellow retangle is 6 an the total number of s that nee to be arrange is 0 it follows that the two numbers we are looking for multipl to 6 an a up to 0. These numbers are an 8 whih an also be seen in () where the onstant prout graph of = 6 an the onstant sum graph of + = 0 intersets at ( 8) an (8 ). a.0 p = 8.00 q =.00 = a( p)( q) = S = P p q ( ) ( 0.00) + (.0 8.0) (0.00) (8.0.0) (a) (b) () Fig. 5 Three visual representations of = ( + )( + 8) = A NEW PATH: LINE AND HYPERBOLA Aroun the same time I starte rethinking first-ear algebra with the help of m olleagues espeiall Anita Wah. In line with the tren to look at man aspets of mathematis in terms of funtions she suggeste that we ask stuents to make a onnetion between the intersetions of the graphs of onstant sums an onstant prouts an the fatoring of trinomials (Wah an Piiotto 994 p. 76.) Over time this iea turne into a stanar assignment (a minireport or poster) for ninth graers at m shool an was epane to inlue the fatore form of quarati funtions an the fatoring of trinomials as represente in Lab Gear (Piiotto 995). (See fig. 5.) Years later when I was orreting a stuent report on these representations it ourre to me that there must be a wa to erive the quarati formula starting with the line an the hperbola. Inee there is. As we saw equation () on the previous page iels an epression for the sum of the roots (p + q = /a). It also gives us an epression for the prout of the roots: pq = /a sine = apq. The onverse of this is that if we have two numbers p an q whose sum is /a an whose prout is /a (with a 0) then p an q are the solutions to the quarati equation a + b + = 0. That the onverse is true an be shown as follows: p + q = b a pq = a ap + aq = b apq = apq + aq = bq apq = () (3) (4) Therefore aq + bq + = 0 whih is what we wante to show. [Eitor s note: Don t worr. We in t lose a root. In step (4) we oul as easil have multiplie eah term b p instea of q. In that ase we woul arrive at ap + bp + = 0.] In other wors solving the first sstem of equations above is equivalent to solving the quarati equation a + b + = 0. If = 0 the equation an reail be solve b fatoring an it is eas to hek that the solution satisfies the quarati formula. If 0 we will solve the sstem in () above with the help of a graph. To make the proess learer we will use the equivalent equations + = b/a an = /a. Vol. 0 No. 6 Februar 008 Mathematis Teaher 475

4 = M = /a N + = /a q /a (p q) p = /a + = /a Fig. 6 The sstem ma have no solutions in ase. Case : /a > 0 In this ase it is possible to have no solution to this sstem (see fig. 6). If there is a solution we have the situation illustrate in figure 7. Consier the square with opposite verties at the origin an M an the square with opposite verties at the origin an N. A solution to our sstem eists if the ifferene of their areas (shae in the figure) is nonnegative. M has oorinates b b a a beause its oorinates satisf N has oorinates N = b + =. a M Fig. 7 Case ma have a solution. a a beause its oorinates satisf = =. a = (p q) = /a + = /a Fig. 8 Determining will provie a solution to the sstem. It follows that the shae area is b b =. a Beause the enominator is alwas positive the fration is nonnegative when its numerator b is nonnegative. We have re-reisovere the isriminant! To solve this sstem we nee to fin p an q in terms of a b an. Beause /a is the sum of p + q /a is the a average. In other wors for some number we have More ompatl we have p = b a q b + an = a. = b ± a an we have solve this sstem if we an fin. To o that onsier figure 8 where is the sie of the shae square. That square has an area equal to the area of the larger square (the one with opposite verties at the origin an M) minus its unshae portion. It is not har to see that this unshae area equals the area of the retangle with opposite verties at the origin an (p q). But beause the hperbola (p q) is the lous of points whose oorinates have a onstant prout this retangle has the same area as the square with opposite verties at the origin an N. Therefore the shae square has the same area as the shae polgon of figure 6 an area that we have alrea alulate. We onlue that b = = b a 476 Mathematis Teaher Vol. 0 No. 6 Februar 008

5 (p q) = a = = /a + = /a + = a N Fig. 9 Case will alwas provie a solution. an therefore that M ( a a) = ± b b a as epete. (p q) Case : /a < 0 In this ase a solution alwas eists (as an be seen in fig. 9). To fin the solution one again note that b = ± a with the sie of the shae square (see fig. 0). To fin the area of the square note that the retangular part below the -ais is ongruent to the retangle immeiatel to its right whih in turn is ongruent b smmetr aross the = line to the retangle sitting atop the shae square. So the area of the square equals the area of its part in the first quarant plus the area of the retangle having opposite verties at the origin an. But that retangle has an area equal to that of the square with opposite verties at N an the origin (fig. ). So b = + a an the erivation proees as above. FURTHER EXPLORATIONS What makes our profession enlessl fasinating is the interpla between our own eploration of mathematis an that of our stuents. This ase is emblemati: I was motivate to seek nontraitional erivations beause of the hallenge of teahing for unerstaning. I enjoe the searh for them eepene m unerstaning an was reware with epane options on how to present a ore part of the high shool mathematis urriulum. M hallenge to reaers of this olumn: Can ou use a similar approah in three imensions to erive Carano s formula for the solution of ubi equations? Fig. 0 As before is the sie of the shae square. The area of the shae square to the left of the -ais is equal to the area of the retangle with opposite verties at the origin an sine both have area equal to /a. Consier the upper part of the retangle with opposite verties at an (0 /a). It is ongruent to its mirror image aross the = line whih is in turn ongruent to the shae retangle below the -ais in figure 0. Therefore the shae areas in figures 0 an are equal. + = a ( a ) N a = a = Fig. Determining the solution when /a < 0 M ( a a) (p q) Vol. 0 No. 6 Februar 008 Mathematis Teaher 477

6 Eitor s notes: We espeiall enjo eamining something ver familiar from an entirel new perspetive. Here to hange perspetive the author swithes variables. The traitional erivation of the quarati formula solves the equation a + b + = 0 (in one variable) b showing how epens on (i.e. is erive from) the values of a b an. If we think of the funtion f() = a + b + then those same values of are the zeros of the funtion. Graphing = f() we see these as the -interepts. But Piiotto presents et another wa to look at the solutions to a + b + = 0. Using p an q to name the two values of that are the solutions Piiotto points out that p an q must satisf two onstraints: one that esribes their sum p + q an one that esribes their prout pq. These two onstraints epress q as a funtion of p in two ifferent was: as a sstem of simultaneous equations onl one of whih is linear. Where the graphs interset both onstraints are met an so the oorinates of either intersetion are the solutions to the original equation a + b + = 0. Piiotto shows how familiar elements of the quarati formula an be erive from ertain features of this graph. The hange of perspetive shows how the two roots of the quarati relate to eah other. Piiotto hallenges the reaer to eten the iea presente here into three imensions. This is a hallenge also pose b Marion Walter in a submission to this epartment that we hope to publish in the future. Her hallenge involves elevating a familiar plane geometr theorem onerning the intersetion of hors of a irle to three imensions. Our invitation to reaers is to eplore other eamples of moving a familiar iea in high shool mathematis from two imensions to three an submit the results to this epartment. REFERENCES Piiotto Henri. Lab Gear Ativities for Algebra. New York: Creative Publiations 995. Wah Anita an Henri Piiotto. Algebra: Themes Tools an Conepts. New York: Creative Publiations HENRI PICCIOTTO math-e@piiotto. org hairs the mathematis epartment at the Urban Shool of San Franiso. A urriulum eveloper an onsultant as well as a teaher he shares ieas on his Web site: Mathematis Teaher Vol. 0 No. 6 Februar 008