Presentazione di PowerPoint

12/07/2006

Equazioni – Disequazioni di 2° grado
Avanti

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12/07/2006

Contenuti

•Forma normale di una e= quazione di secondo grado =

•Equazioni di secondo grado incomplete - complete

•Risoluzione di una equazione incompleta - completa

•Condizioni sul discriminante

•Risoluzione di una equazione di 2° gr= ado- metodo grafico

•Soluzioni di un’equazione di 2° grado e scomposizione in fattori&= #13;

•Disequazioni di 2° grado <= /span>

•Risoluzione di una disequazioni di 2° grado

•Disequazioni di gr= ado superiore al secondo

•Disequazioni frazionarie

Avanti

Indietro

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Forma normale

•    Un’equazione di secondo grado, in forma normale, è del tipo        = ;

•        = ;            &n= bsp;            ax2 + bx + c =3D 0&#= 13;

•è ordinata in ordine decrescente rispetto le potenze della incognita

•non pres= enta termini simili <= /nobr>

•

•   &nbs= p; Esempio

•     4x= 2 - 5x - 1 =3D 0

Avanti

Indietro

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12/07/2006
<= v:shapetype id=3D"_x0000_t93" coordsize=3D"21600,21600" o:spt=3D"93" adj=3D"16200,540= 0" path=3D"m@0,l@0@1,3375@1,3375@2@0@2@0,21600,21600,10800xem1350@1l1350@2,= 2700@2,2700@1xem0@1l0@2,675@2,675@1xe">
Esempio

•    Scrivere l’equazione : 3= x= (2 - x) =3D 2(x + 1) in forma norma= le

•si eseguono i calcoli      6x - 3x2 =3D 2x +2 =

• =

•si applica il 1°principio d’equivalenza

•        = ;            &n= bsp;            = ;          6x – 3x2 - 2x - 2 =3D 0&= #13;

•si sommano i termini simili
        = ;            &n= bsp;            = ;     –3x2 + 4x –2 =3D 0

•si applica il 2°principio di equivalenza

•        = ;            &n= bsp;            = ;          3x2 - 4= x= + 2 =3D= 0

Avanti

Indietro

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12/07/2006

Equazioni incomplete

•       Un’equazione di secondo grado di dice incompleta se manca di= uno o più termini:

•       = 3;

n       se<= span style=3D'mso-spacerun:yes'> b =3D 0 si ha   ax2 + c =3D 0 e l’equazione si dice pura

n       se<= span style=3D'mso-spacerun:yes'> c =3D 0 si ha ax2 + bx =3D 0 e l’equazione si dice spuria

Avanti

Indietro

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12/07/2006

nax2 + c =3D 0

§Applicando i princ= ipi di equivalenza si ha

§<= /span>

n        = ;

•se -c/= a= > 0 = allora l’equazione ammette due soluzioni reali e opposte

•        = ;

•

•se -c/= a= =3D 0, cioè c =3D 0, allora l’equazione ammette = due soluzioni nulle: in tal caso l’equazione si dice monomia =

•se -c/a < 0 allora l’equazione non ammette soluzioni reali

=

Equazione pura

Avanti

Indietro

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12/07/2006

•Equazione spuria

ax= 2 + bx =3D 0 = ;

§<= /span> Si scompone in fattori =

x<= /span>(ax + b) =3D 0

§<= /span> Si applica la legge di annullament= o del prodotto

x<= /span> =3D 0 o ax + b =3D 0

§<= /span> Si ottengono due soluzioni reali d= i cui una nulla

        = ;

Avanti

Indietro

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12/07/2006

Equazione completa

•    Per risolvere un’equazione di 2° grado completa

• ax2 + bx + c =3D 0

•     si utilizza la formula

•

•        = ;

• =

•   &nbs= p; La quantità sotto radice b2 – 4ac si chia= ma discriminante e si indica con la lettera D (leggi “delta”). Tale quantità “discrimina” i vari casi.

Avanti

Indietro

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12/07/2006

Condizioni sul discriminante

= •

•Se D > 0 si hanno due soluzioni reali e distinte

= •
•Se D =3D 0 si hanno due soluzioni reali e coincidenti (cioè una soluzione doppia)
= •
•Se D < 0 non si hanno soluzioni reali e l’equazion= e si dice impossibile in R&#= 13;

Avanti

Indietro

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12/07/2006

Esempio 1

•   &nbs= p; Risolvere l’equazione &= nbsp; (x - 1)(3x= + 1) = =3D 2x= = - 1

•

•si riduce in forma normale, ottenendo l’equazione spuria

•3x2 - 4x= =3D 0

•si raccoglie x        = ;    x(3x - 4) =3D 0

•si applica la legge di annullamento del prodotto ponendo

•x =3D 0 Ú 3x - 4 =3D 0=

•si ricavano due soluzioni reali

•x =3D 0 Ú = = x= =3D 4/3=

Avanti

Indietro

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12/07/2006

Esempio 2

•     Risolvere l’equazione     x(x - 1) =3D <= /span>- x(x + 1) + 8

<= span style=3D'font-size:75%;color:#666699;mso-color-index:7;visibility:hidden'>= •
•<= /span>eseguend= o i calcoli e applicando il 1&de= g; principio di equivalenza, si ottiene l’equazione pura
•2x2 - 8 =3D 0

<= span style=3D'font-size:75%;color:#666699;mso-color-index:7'>•l’equazione ammette due soluzioni reali e opposte

•x2 =3D 4

•x =3D ± = 2=

Avanti

Indietro

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12/07/2006

Esempio 3

•   &nbs= p; Risolvere l’equazione frazionaria

•

•

•si applica il 2°principio d’equivalenza, ponendo le condizio= ni        = ;            &n= bsp; , (*)

•si riduce in forma normale, ottenendo l’equazione completa

•poiché D > 0 = si hanno due soluzioni reali e distinte

• =

• =

•la soluzione x =3D1 no= n è accettabile per le (*)

Avanti

Indietro

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12/07/2006

Risoluzione di un’equazione di 2° grado
Metodo grafico

•     L’equazio= ne ax2 + bx + c =3D 0 s= i ottiene dall’intersezione della parabola di equazione y =3D ax2 + bx + c con l= 217;asse delle ascisse, la cui equazione è y =3D 0. =
•     Esempio

•     Sia x2 - 4x= =3D 0&#= 13;

•D > 0

•l’= equazione ha due soluzioni

•     reali e distint= e x =3D= 0= = o= x =3D 4 &= #13;

•la parab= ola di equazione y =3D x2 - 4x ha perta= nto due intersezioni con l’asse delle ascisse

Avanti

Indietro

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12/07/2006

Soluzioni di un’equazione <= br> di 2° grado e scomposizione in fattori

•       Data l’equazione ax2 + bx + c =3D 0, si dimostra che essa è equivalente a

•a(= x= –= x= 1)(x – x= 2) =3D 0

•       dov= e x= 1 ed x2 sono le soluzioni dell’equaz= ione determinate applicando la formula risolutiva delle equazioni complete di 2° grado.

•       Esempi =

•        1. L’equazione = 4x= 2 - 3x - 1 =3D 0 ha soluzioni x1 =3D -1/4 e

•        = ;      x2=3D 1 e pertanto &= egrave; equivalente a 4(x + ¼) (x – 1) =3D 0.

•        2. Viceversa se l’equazione si presenta nella forma <= /div>
•        = ;    3(x + 2)(x – = 5) =3D 0, le soluzioni sono x1 =3D –2 e x= 2=3D 5.

Avanti

Indietro

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12/07/2006

Disequazioni di 2° grado

•    Sono scritte nella forma:

•ax2 + bx + c < 0<= span style=3D'mso-spacerun:yes'> (oppure ax2 + bx + c £ 0)

= •

•ax2 + bx + c > 0<= span style=3D'mso-spacerun:yes'> (oppure ax2 + bx + c ³ 0)

= •
di x= il trin= omio ax2 + bx + c è negativo, nullo o positi= vo, cio&egra= ve; per quali valori di x la parabola di equazione
•y =3D ax2 + bx + c

•   &nbs= p; è negativa, nulla o positiva.

Avanti

Indietro

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12/07/2006

Risoluzione di una
disequazione di 2° grado

•       Se la disequazione è del tipo

• ax2 + bx + c > 0 (oppure ax2 + bx + c ³ 0) con a > 0

•        si possono presentare tre casi (D > 0, D =3D 0, D < 0).

•        Prima di passare all’analisi di questi tre casi si ricorda ch= e se a > 0 = la parabola di equazione y =3D ax2 + bx + c ha la concavità rivolta verso l’alto e che se a < 0 = è sempre possibile cambiare il segno a tutti i termini della disequazione (purchè se ne cambi il verso!!) e ricondursi ancora al caso = a= > 0.=

Avanti

Indietro

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12/07/2006

Risoluzione di una
disequazione di 2° grado

Ø       ax2 + bx + c > 0 (= ax2 + bx + c ³ 0) con&nb= sp; a > 0

Ø =

&Osl= ash;se D >= ; 0 l’equazione associata ax2 + bx + c =3D 0 ha= due soluzioni reali e distinte x1 e x2, quindi la parabola di equazione y =3D ax2 + bx + c interseca l’asse delle ascisse in due punti distinti ed è positiva (positiva o nulla)   se   x < x1 Ú x > x2   (x = £ x1 &Uacut= e; = x= = ³ x2)

Avanti

Indietro

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12/07/2006

Risoluzione di una
disequazione di 2° grado

Øax2 + bx + c > 0 (= ax2 + bx + c ³ 0) con&nb= sp; a > 0

Ø =

&Osla= sh;se D =3D= 0 l’equazione associata ax2 + bx + c =3D 0 ha= due soluzioni reali e coincidenti x1 =3D x2 , quindi= la parabola di equazione y =3D ax2 + bx + c interseca l’asse delle ascisse in un solo punto (è tangente) ed è positiva (positiva o nulla) per ogni valore di x con= x ¹ x1   (per ogni valore di= x= )=

Avanti

Indietro

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charset="us-ascii" Presentazione di PowerPoint

12/07/2006

Risoluzione di una
disequazione di 2° grado

Øax2 + bx + c > 0 (= ax2 + bx + c ³ 0) con&nb= sp; a > 0

Ø =

Øse D < 0 l’equazione associata ax2 + bx + c =3D 0 non ha soluzioni reali,   quindi la parabola di equazione= y= =3D ax2 + bx + c non interseca l’asse delle ascisse ed è positiva (<= span lang=3DIT style=3D'font-size:75%;color:#3694B6;mso-color-index:5'>positiva= o nulla) per ogni valore di x (per ogni valore di x)

Avanti

Indietro

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12/07/2006

Esempio 1

•   &nbs= p; Risolvere la disequazione       x= 2 + 4x + 4 > 0

•la parabola di equazione y =3D x= 2 + 4x + 4 ha la concavità = rivolta = verso l’alto

•l’equazione, associata alla disequazione, x2 + 4x + 4 =3D 0 = ha D =3D 0 e soluzione doppia x =3D -2

•la parabola è allora tangente all’asse x

•   &nbs= p; nel vertice di ascissa -2

•la parabola &egrav= e; pertanto positiva per ogni

•   &nbs= p; valore di x ad esclusione del valore
•   &= nbsp; x =3D -2 per cui è nulla

•La soluzione è= " = x= = ¹ -2

Avanti

Indietro

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12/07/2006

Esempio 2

     Risolvere la disequazione      x2 + 2x + 4 > 0        = ;

•    la parabola di equazione y = =3D x2 + 2x + 4 ha la concavità

      rivolta v= erso l’alto

•    l’equazione, associata alla disequazione, x2 + 2x + 4 =3D 0

     ha D < 0 e pertanto non ha soluzioni reali

•    la parabola non inter= seca l’asse x

     ed è per= tanto positiva per

     ogni valore di = x

•    la soluzione <= span lang=3DIT style=3D'layout-flow:horizontal;color:#666699;mso-color-index:7'= >è " x <= span lang=3DIT style=3D'font-family:Symbol;layout-flow:horizontal;color:#666699; mso-color-index:7;mso-char-type:symbol'>Î <= span lang=3DIT style=3D'layout-flow:horizontal;color:#666699;mso-color-index:7'= >R

Avanti

Indietro

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12/07/2006

Esempio 3

•         Risolvere

• =

• =

•        Risolvere un sistema di disequazioni significa determinare le soluz= ioni che esse hanno in comune

•        = ; S1:

•        = ; S2:        = ;            &n= bsp;            = ;            &n= bsp;            = ;

•        = ; S3:        = ;            &n= bsp;            = ;            &n= bsp;            = ;            &n= bsp;

•

•        = ; Soluzione del sistema:

= =
Avanti

Indietro

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12/07/2006

Disequazioni di grado
superiore al secondo

•     Per risolvere u= na disequazione di grado superiore al secondo può essere conveniente, = con una fattorizzazione, scriver= le nella forma

•A(x)= ×B(x)= ×C(x) > 0= (oppure ³)

•A(x)= ×B(x)= ×C(x) < 0= (oppure £)

• In entrambi i cas= i, per risolvere:

•si pone = ogni fattore > 0 (oppure ³= 0)=

•si risol= ve ciascuna disequazione e si rappresentano graficamente le soluzioni

•la soluz= ione finale è data dagli intervalli in cui il prodotto dei segni è positivo o negativo (oppure positivo o nullo, negativo o nullo).

Avanti

Indietro

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12/07/2006

Esempio

•   &nbs= p; Risolvere la disequazione

•<= /span>

•si scompone in fattori

• =

•si pone ogni fattore > 0 e si risolve ciascuna disequazione

•

•   &nbs= p; 1° Fattore:

•

•   &nbs= p; 2° Fattore:

•

•   &nbs= p; 3° Fattore:

Avanti

Indietro

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12/07/2006

Segue l’esempio =

•si rappresentano graficamente le soluzioni

•        = ;            &n= bsp;            = ;

• =

• =

• =

• =

• =
•la soluz= ione della disequazione è data dagli intervalli in cui il p= rodotto dei segni è positivo
•

Avanti

Indietro

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Presentazione di PowerPoint

12/07/2006

Disequazioni frazionarie

•     Sono scritte ne= lla forma

•        = ;            &n= bsp;            (oppure ³)        = ;             (oppure = £)

• =
compare al denomin= atore. In entrambi i casi, per risolvere
•si pone il numerat= ore > 0 (oppure ³ 0)

•si pone il denomin= atore > 0
•si risol= ve ciascuna disequazione e si rappresentano graficamente le soluzioni
•la soluzione finale è data dagli intervalli in cui il prodotto dei segni è posit= ivo o negativo (oppure positivo o nullo, negativo o nullo).

Avanti

Indietro

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12/07/2006

Esempio

•   &nbs= p; Risolvere la disequazione

•<= /span>

•    In tal caso, poich&ea= cute; il numeratore è scomposto in fattori, si pu&ograv= e; procedere ponendo ogni fattore > 0

•

•    1° Fattore:

•

•    2° Fattore:

•

•    3° Fattore:

Avanti

Indietro

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12/07/2006

Segue l’esempio =

•si rappresentano graficamente le soluzioni

•        = ;            &n= bsp;            = ;

• =

• =

• =

• =

• =
•la soluz= ione della disequazione è data dagli intervalli in cui il p= rodotto dei segni è negativo
•

Indietro

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loop ){ href=3D unescape(href); if( window.event ) window.event.cancelBubble =3D true; pptSound.src =3D href; pptSound.loop =3D loop ? -1 : 0; } function DocumentOnClick() { if( IsNts() || parent.HideMenu() ) return; if( ( g_allowAdvOnClick && !parent.IsFramesMode() ) || (event && (event.keyCode=3D=3D32) ) ) parent.GoToNextSld(); } var g_supportsPPTHTML =3D SupportsPPTHTML(), g_scaleInFrame =3D 1, gId=3D""= , g_bgSound=3D"", g_scaleHyperlinks =3D false, g_allowAdvOnClick =3D 1, g_showInBrowser = =3D 0, gLoopCont =3D 1, gUseSldTimings =3D 0; var g_showAnimation =3D g_supportsPPTHTML && SupportsPPTAnimation() && ( (w= indow.name=3D=3D"PPTSld" && !parent.IsFramesMode()) || g_showInBrowser );va= r g_animManager =3D null; var g_animUseRuntime =3D false; var g_animItemsToHide, g_animInteractiveItems, g_animSlideTime; var g_animMainSequence =3D null; var ENDSHOW_MESG=3D"Fine della presentazione. Fare clic per uscire.", SCREE= N_MODE=3D"Frames", gIsEndShow=3D0, NUM_VIS_SLDS=3D28, SCRIPT_HREF=3D"script= .js", FULLSCR_HREF=3D"fullscreen.htm"; var gCurSld =3D gPrevSld =3D 1, g_offset =3D 0, gNtsOpen =3D gHasNts =3D gO= tlTxtExp =3D 0, gHasNarration =3D 0, gOtlOpen =3D true window.gPPTHTML=3DSupportsPPTHTML() var gMainDoc=3Dnew Array(new hrefList("slide0013.htm",1,-1,1),new hrefList(= "slide0003.htm",1,-1,1),new hrefList("slide0014.htm",1,-1,1),new hrefList("= slide0015.htm",1,-1,1),new hrefList("slide0010.htm",1,-1,1),new hrefList("s= lide0016.htm",1,-1,1),new hrefList("slide0012.htm",1,-1,1),new hrefList("sl= ide0007.htm",1,-1,1),new hrefList("slide0008.htm",1,-1,1),new hrefList("sli= de0018.htm",1,-1,1),new hrefList("slide0019.htm",1,-1,1),new hrefList("slid= e0020.htm",1,-1,1),new hrefList("slide0021.htm",1,-1,1),new hrefList("slide= 0009.htm",1,-1,1),new hrefList("slide0026.htm",1,-1,1),new hrefList("slide0= 023.htm",1,-1,1),new hrefList("slide0039.htm",1,-1,1),new hrefList("slide00= 37.htm",1,-1,1),new hrefList("slide0038.htm",1,-1,1),new hrefList("slide002= 7.htm",1,-1,1),new hrefList("slide0028.htm",1,-1,1),new hrefList("slide0030= .htm",1,-1,1),new hrefList("slide0031.htm",1,-1,1),new hrefList("slide0032.= htm",1,-1,1),new hrefList("slide0033.htm",1,-1,1),new hrefList("slide0034.h= tm",1,-1,1),new hrefList("slide0035.htm",1,-1,1),new hrefList("slide0040.ht= m",1,-1,1)); function FullScrInit() { g_allowAdvOnClick =3D GetCurSld().mAdvOnClk document.body.style.backgroundColor=3D"black" document.oncontextmenu=3Dparent._CM; document.onkeydown =3D _KDH; document.ondragstart=3DCancel document.onselectstart=3DCancel self.focus() } function Redirect( frmId ) { var str=3Ddocument.location.hash,idx=3Dstr.indexOf('#'), sId=3DGetSldId() if(idx>=3D0) str=3Dstr.substr(1); if( window.name !=3D frmId && ( sId !=3D str) ) { obj =3D GetObj("Main-File") window.location.href=3Dobj.href+"#"+sId return 1 } return 0 } var MHTMLPrefix =3D CalculateMHTMLPrefix(); function CalculateMHTMLPrefix() { if ( document.location.protocol =3D=3D 'mhtml:') { href=3Dnew String(document.location.href) Start=3Dhref.indexOf('!')+1 End=3Dhref.lastIndexOf('/')+1 if (End < Start) return href.substring(0, Start) else return href.substring(0, End) } return ''; } function GetTags(base,tag) { if(g_supportsPPTHTML) return base.all.tags(tag); else return base.getElementsByTagName(tag); } function UpdNtsPane(){ if(frames["PPTNts"]) PPTNts.location.replace( MHTMLP= refix+GetHrefObj( gCurSld ).mNtsHref ) } function UpdNavPane( sldIndex ){ if(gNavLoaded) PPTNav.UpdNav() } function UpdOtNavPane(){ if(gOtlNavLoaded) PPTOtlNav.UpdOtlNav() } function UpdOtlPane(){ if(gOtlLoaded) PPTOtl.UpdOtl() } function SetHasNts( fVal ) { if( gHasNts !=3D fVal ) { gHasNts=3DfVal UpdNavPane() } } function ToggleOtlText() { gOtlTxtExp=3D!gOtlTxtExp UpdOtlPane() } function ToggleOtlPane() { frmset=3Ddocument.all("PPTHorizAdjust") frm=3Ddocument.all("PPTOtl") if( gOtlOpen ) frmset.cols=3D"*,100%" else frmset.cols=3D"25%,*" gOtlOpen=3D!gOtlOpen frm.noResize=3D!frm.noResize UpdOtNavPane() } function ToggleNtsPane() { frmset=3Ddocument.all("PPTVertAdjust") frm=3Ddocument.all("PPTNts") if( gNtsOpen ) frmset.rows=3D"100%,*" else frmset.rows=3D"*,20%" gNtsOpen=3D!gNtsOpen UpdNtsPane() } function ClearMedia() { if (PPTSld.pptSound) PPTSld.pptSound.loop =3D 0; } function FullScreen() { if ( PPTSld.g_animUseRuntime ) PPTSld.document.body.pause(); ClearMedia(); var href =3D ( document.location.protocol =3D=3D 'mhtml:') ? FULLSCR_HREF = : FULLSCR_HREF+"#"+GetHrefObj(gCurSld).mSldHref; if(PPTNav.event.ctrlKey) { var w =3D (window.screen.availWidth * 1.0) / 2.0 var h =3D w * (PPTSld.g_origH * 1.0) / PPTSld.g_origW win =3D window.open( MHTMLPrefix+href,null,"toolbar=3D0,resizable=3D1,top= =3D0,left=3D0," + "width=3D"+ w + ",height=3D" + h ); if( PPTSld.g_animUseRuntime ) win.document.body.PPTSldFrameset=3Dwindow; } else { win =3D window.open( MHTMLPrefix+href,null,"fullscreen=3Dyes" ); if( PPTSld.g_animUseRuntime ) win.document.body.PPTSldFrameset=3Dwindow; } } function ToggleVNarration() { rObj=3DPPTSld.document.all("NSPlay") if( rObj && !PPTSld.g_animUseRuntime ) { if( (rObj.playState =3D=3D 1)||(rObj.playState =3D=3D 0) ) rObj.Play() else if( rObj.playState =3D=3D 2 ) rObj.Pause() else return; } else if( PPTSld.g_animUseRuntime ) { narObj =3D PPTSld.document.all("narrationID") if( narObj ) narObj.togglePause() } } function GetCurSldNum() { obj=3DGetHrefObj(gCurSld) if( obj.mOrigVis =3D=3D 1 ) return obj.mSldIdx else return gCurSld } function GetNumSlds() { if( GetHrefObj(gCurSld).mOrigVis =3D=3D 1 ) return GetSldList().mNumVisSlds; else return GetSldList().mList.length } function GetSldNum( href ) { for(ii=3D0; ii 1 ) PopSldList(); else if( !IsFramesMode() ) { if( gLoopCont ) GoToFirst() else EndShow() } } function GoToPrevSld() { ii=3DgCurSld-1 if( ii > 0 ) { obj=3DGetHrefObj(ii) while ( obj && ( obj.mVis =3D=3D 0 ) && ( ii>0 ) ) obj=3DGetHrefObj(--ii) if( ii =3D=3D 0 ) ii=3D1 GoToSldNum(ii) } } function GoToFirst(){ GoToSld( GetHrefObj(1).mSldHref ) } function GoToLast() { ii=3DGetSldList().mList.length if( ii !=3D gCurSld ) GoToSld( GetHrefObj(ii).mSldHref ) } function GoToSldNum( num ) { if( PPTSld.event ) PPTSld.event.cancelBubble=3Dtrue obj =3D GetHrefObj( num ) obj.mVis=3D1 gPrevSld=3DgCurSld gCurSld =3D num; PPTSld.location.replace(MHTMLPrefix+obj.mSldHref) if( IsFramesMode() ) { UpdNavPane(); UpdOtlPane(); UpdNtsPane() } } function GoToSld( href ) { if( PPTSld.event ) PPTSld.event.cancelBubble=3Dtrue GetHrefObj( GetSldNum(href) ).mVis=3D1 PPTSld.location.replace(MHTMLPrefix+href) } function SldUpdated( id ) { if( id =3D=3D GetHrefObj(gCurSld).mSldHref ) return gPrevSld=3DgCurSld gCurSld=3DGetSldNum(id) if( IsFramesMode() ) { UpdNavPane(); UpdOtlPane(); UpdNtsPane() } } function PrevSldViewed(){ GoToSld( GetHrefObj(gPrevSld).mSldHref ) } function HasPrevSld() { return ( gIsEndShow || ( gCurSld !=3D 1 && GetHrefO= bj( gCurSld-1 ).mVis =3D=3D 1 )||( GetCurSldNum() > 1 ) ) } function HasNextSld() { return (GetCurSldNum() !=3D GetNumSlds()) } function CloseWindow() { if( HideMenu() ) return; var event =3D PPTSld.event; if( !IsFramesMode() && event && (event.keyCode=3D=3D27 || event.keyCode=3D= =3D32 || event.type=3D=3D"click" ) ) window.close( self ); CatchNumKeys( self, event ); } function Unload() { gIsEndShow=3D0; } function SetupEndShow() { gIsEndShow=3D1; PPTSld.document.body.scroll=3D"no"; PPTSld.document.onkeypress=3DCloseWindow; PPTSld.document.onclick=3DCloseWindow; PPTSld.document.oncontextmenu=3D_CM; } function EndShow() { if( IsFramesMode() ) return if( PPTSld.event ) PPTSld.event.cancelBubble=3Dtrue doc=3DPPTSld.document doc.open() doc.writeln('

' + ENDSHOW_MESG + '
') doc.close() } function SetSldVisited(){ GetSldList().mList[gCurSld-1].mVisited=3Dtrue } function IsSldVisited(){ return GetSldList().mList[gCurSld-1].mVisited } function hrefList( sldHref, visible, advDelay, advClk ) { this.mSldHref=3D this.mNtsHref =3D sldHref this.mOrigVis=3D this.mVis =3D visible this.mVisited=3D false this.mAdvDelay=3D advDelay this.mAdvOnClk=3D advClk } function SldList(arr,curSld,fEnd) { this.mCurSld =3D curSld; this.mList =3D new Array(); var idx =3D 1; for(ii=3D0;ii 0) { PushSldList(sldList,fEnd); gCurSld =3D 1; } else if( PPTSld.event ) PPTSld.event.cancelBubble=3Dtrue } function PushSldList(arr,fEnd) { var ii =3D gSldStack.length; gSldStack[ii] =3D new SldList(arr,gCurSld,fEnd); GoToSld( gSldStack[ii].mList[0].mSldHref ); } function PopSldList() { if (gSldStack[gSldStack.length-1].fEndShow) EndShow() else { gCurSld =3D gSldStack[gSldStack.length-1].mCurSld; gSldStack[gSldStack.length-1] =3D null; gSldStack.length--; var sldList =3D gSldStack[gSldStack.length-1]; GoToSld( sldList.mList[gCurSld - 1].mSldHref ); } } var custShowList=3Dnew Array(); function ImgBtn( oId,bId,w,action ) { var t=3Dthis t.Perform =3D _IBP t.SetActive =3D _IBSetA t.SetInactive=3D _IBSetI t.SetPressed =3D _IBSetP t.SetDisabled=3D _IBSetD t.Enabled =3D _IBSetE t.ChangeIcon =3D null t.UserAction =3D action t.ChgState =3D _IBUI t.mObjId =3D oId t.mBorderId=3D bId t.mWidth =3D w t.mIsOn =3D t.mCurState =3D 0 } function _IBSetA() { if( this.mIsOn ) { obj=3Dthis.ChgState( gHiliteClr,gShadowClr,2 ) obj.style.posTop=3D0 } } function _IBSetI() { if( this.mIsOn ) { obj=3Dthis.ChgState( gFaceClr,gFaceClr,1 ) obj.style.posTop=3D0 } } function _IBSetP() { if( this.mIsOn ) { obj=3Dthis.ChgState( gShadowClr,gHiliteClr,2 ) obj.style.posLeft+=3D1; obj.style.posTop+=3D1 } } function _IBSetD() { obj=3Dthis.ChgState( gFaceClr,gFaceClr,0 ) obj.style.posTop=3D0 } function _IBSetE( state ) { var t=3Dthis GetObj( t.mBorderId ).style.visibility=3D"visible" if( state !=3D t.mIsOn ) { t.mIsOn=3Dstate if( state ) t.SetInactive() else t.SetDisabled() } } function _IBP() { var t=3Dthis if( t.mIsOn ) { if( t.UserAction !=3D null ) t.UserAction() if( t.ChangeIcon ) { obj=3DGetObj(t.mObjId) if( t.ChangeIcon() ) obj.style.posLeft=3Dobj.style.posLeft+(t.mCurState-4)*t.mWidth else obj.style.posLeft=3Dobj.style.posLeft+(t.mCurState-0)*t.mWidth } t.SetActive() } } function _IBUI( clr1,clr2,nextState ) { var t=3Dthis SetBorder( GetObj( t.mBorderId ),clr1,clr2 ) obj=3DGetObj( t.mObjId ) obj.style.posLeft=3Dobj.style.posLeft+(t.mCurState-nextState)*t.mWidth-obj= .style.posTop t.mCurState=3DnextState return obj } function TxtBtn( oId,oeId,action,chkState ) { var t=3Dthis t.Perform =3D _TBP t.SetActive =3D _TBSetA t.SetInactive=3D _TBSetI t.SetPressed =3D _TBSetP t.SetDisabled=3D _TBSetD t.SetEnabled =3D _TBSetE t.GetState =3D chkState t.UserAction =3D action t.ChgState =3D _TBUI t.mObjId =3D oId t.m_elementsId=3D oeId t.mIsOn =3D 1 } function _TBSetA() { var t=3Dthis if( t.mIsOn && !t.GetState() ) t.ChgState( gHiliteClr,gShadowClr,0,0 ) } function _TBSetI() { var t=3Dthis if( t.mIsOn && !t.GetState() ) t.ChgState( gFaceClr,gFaceClr,0,0 ) } function _TBSetP() { if( this.mIsOn ) this.ChgState( gShadowClr,gHiliteClr,1,1 ) } function _TBSetD() { this.ChgState( gFaceClr,gFaceClr,0,0 ) this.mIsOn =3D 0 } function _TBSetE() { var t=3Dthis if( !t.GetState() ) t.ChgState( gFaceClr,gFaceClr,0,0 ) else t.ChgState( gShadowClr,gHiliteClr,1,1 ) t.mIsOn =3D 1 } function _TBP() { var t=3Dthis if( t.mIsOn ) { if( t.UserAction !=3D null ) t.UserAction() if( !t.GetState ) return if( t.GetState() ) t.SetPressed() else t.SetActive() } } function _TBUI( clr1,clr2,lOffset,tOffset ) { SetBorder( GetObj( this.mObjId ),clr1,clr2 ) Offset( GetObj( this.m_elementsId ),lOffset,tOffset ) } function Offset( obj, top, left ){ obj.style.top=3Dtop; obj.style.left=3Dle= ft } function SetBorder( obj, upperLeft, lowerRight ) { s=3Dobj.style; s.borderStyle =3D "solid" s.borderWidth =3D 1 s.borderLeftColor =3D s.borderTopColor =3D upperLeft s.borderBottomColor=3D s.borderRightColor =3D lowerRight } function GetBtnObj(){ return gBtnArr[window.event.srcElement.id] } function BtnOnOver(){ b=3DGetBtnObj(); if( b !=3D null ) b.SetActive() } function BtnOnDown(){ b=3DGetBtnObj(); if( b !=3D null ) b.SetPressed() } function BtnOnOut(){ b=3DGetBtnObj(); if( b !=3D null ) b.SetInactive() } function BtnOnUp() { b=3DGetBtnObj() if( b !=3D null ) b.Perform() else Upd() } function GetNtsState(){ return parent.gNtsOpen } function GetOtlState(){ return parent.gOtlOpen } function GetOtlTxtState(){ return parent.gOtlTxtExp } function NtsBtnSetFlag( fVal ) { s=3Ddocument.all.item( this.m_flagId ).style s.display=3D"none" if( fVal ) s.display=3D"" else s.display=3D"none" } function _BSetA_Border(){ b =3D gBtnArr[this.mObjId]; if( b !=3D null ) b.S= etActive() } function _BSetI_Border(){ b =3D gBtnArr[this.mObjId]; if( b !=3D null ) b.S= etInactive() } var gHiliteClr=3D"THREEDHIGHLIGHT",gShadowClr=3D"THREEDSHADOW",gFaceClr=3D"= THREEDFACE" var gBtnArr =3D new Array() gBtnArr["nb_otl"] =3D new TxtBtn( "nb_otl","nb_otlElem",parent.ToggleOtlPan= e,GetOtlState ) gBtnArr["nb_otlElem"] =3D new TxtBtn( "nb_otl","nb_otlElem",parent.ToggleOt= lPane,GetOtlState ) gBtnArr["nb_nts"] =3D new TxtBtn( "nb_nts","nb_ntsElem",parent.ToggleNtsPan= e,GetNtsState ) gBtnArr["nb_prev"]=3D new ImgBtn( "nb_prev","nb_prevBorder",30,parent.GoToP= revSld ) gBtnArr["nb_next"]=3D new ImgBtn( "nb_next","nb_nextBorder",30,parent.GoToN= extSld ) gBtnArr["nb_sldshw"]=3D new ImgBtn( "nb_sldshw","nb_sldshwBorder",18,parent= .FullScreen ) gBtnArr["nb_sldshwBorder"] =3D new TxtBtn( "nb_sldshw","nb_sldshwBorder",pa= rent.FullScreen,null ) gBtnArr["nb_sldshwBorder"].SetActive =3D _BSetA_Border; gBtnArr["nb_sldshwBorder"].SetInactive =3D _BSetI_Border; gBtnArr["nb_sldshwText"] =3D new TxtBtn( "nb_sldshw","nb_sldshwText",parent= .FullScreen,null ) gBtnArr["nb_sldshwText"].SetActive =3D _BSetA_Border; gBtnArr["nb_sldshwText"].SetInactive =3D _BSetI_Border; gBtnArr["nb_voice"] =3D new ImgBtn( "nb_voice","nb_voiceBorder",18,parent.= ToggleVNarration ) gBtnArr["nb_otlTxt"]=3D new ImgBtn( "nb_otlTxt","nb_otlTxtBorder",23,parent= .ToggleOtlText ) gBtnArr["nb_nts"].m_flagId=3D "notes_flag" gBtnArr["nb_nts"].SetFlag =3D NtsBtnSetFlag gBtnArr["nb_otlTxt"].ChangeIcon=3D GetOtlTxtState var sNext=3D"Successivo",sPrev=3D"Precedente",sEnd=3D"Fine presentazione",s= Font=3D"Arial",sArrow=3D"Freccia",sFreeform=3D"Figura a mano libera",sRect= =3D"Rettangolo",sOval=3D"Ovale" function ShowMenu() { BuildMenu(); var doc=3DPPTSld.document.body,x=3DPPTSld.event.clientX+doc.scrollLeft,y= =3DPPTSld.event.clientY+doc.scrollTop m =3D PPTSld.document.all.item("ctxtmenu") m.style.pixelLeft=3Dx if( (x+m.scrollWidth > doc.clientWidth)&&(x-m.scrollWidth > 0) ) m.style.pixelLeft=3Dx-m.scrollWidth m.style.pixelTop=3Dy if( (y+m.scrollHeight > doc.clientHeight)&&(y-m.scrollHeight > 0) ) m.style.pixelTop=3Dy-m.scrollHeight m.style.display=3D"" } function _CM() { if( !parent.IsFullScrMode() ) return; 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Note

Presentaz= ione

Struttura

1

Equazioni – Disequazioni di 2° grad= o

2

Contenuti

Forma normale di una equazione di secondo grado

Equazioni di secondo grado incomplete - complete

Risoluzione di una equazione incompleta - completa

Condizioni sul discriminante

Risoluzione di una equazione di 2° grado- metodo grafico

Soluzioni di un’equazione di 2° grado e scomposizione in fattori

Disequazioni di 2° grado

Risoluzione di una disequazioni di 2° grado

Disequazioni di grado superiore al secondo

Disequazioni frazionarie

3

Forma normale

    Un’equazione di secondo grado, in forma normale, &egrav= e; del tipo

       = ;            &n= bsp;             ax²= + bx + c =3D 0

è ordinata in ordine decrescente rispetto le potenze della incognita

non presenta termini simili

     Ese= mpio

     4x<= sup>2 - 5x - 1 =3D 0

4

Esempio

    Scrivere l’equazione : 3x= (2 - x) =3D 2(x + 1) in forma normale

si eseguono i calcoli      6x = - 3x² =3D 2x +2

si applica il 1°principio d’equivalenza

       = ;            &n= bsp;            = ;           6x – 3x²- 2x - 2 =3D 0

si sommano i termini simili
       = ;            &n= bsp;            = ;      –3x²+ 4x –2 =3D 0

si applica il 2°principio di equivalenza

       = ;            &n= bsp;            = ;           3x²- 4x + 2 =3D 0

5

Equazioni incomp= lete

       Un’equazione di secondo grado di dice incompleta se man= ca di uno o più termini:

       se b =3D 0 si h= a   ax² + c = =3D 0 e l’equazione si dice pura

       se c =3D 0 si h= a ax² + bx =3D 0 e l’equazione si dice spuria

6

"ax^{2 + c
=3D 0"}

ax² + c =3D 0

Applicando i principi di equivalenza si ha

se -c/a > 0 allora l’equazione ammette due soluzioni reali e opposte

se -c/a =3D 0, cio&egr= ave; c =3D 0, allora l’equazione ammette due soluzioni nulle: in tal caso l’equazione si dice monomia

se -c/a < 0 allora l’equazione non ammette soluzioni reali<= /li>

7

"Equazione = spuria"

Equazione spuria

8

Equazione comple= ta

    Per risol= vere un’equazione di 2° grado completa

ax²+ bx + = c =3D 0

     si utilizza la formula

     La quantità sotto radice b² – 4ac si chiama discriminante e si indica con la lettera D (leggi “delta”). Tale quantità “discrimina” i vari casi.

9

Condizioni sul discriminante

Se D > = 0 si hanno due soluzioni reali e distinte

Se D =3D 0= si hanno due soluzioni reali e coincidenti (cioè una soluzione doppia)=

Se D < = 0 non si hanno soluzioni reali e l’equazione si dice impossibile in R

10

Esempio 1

     Ris= olvere l’equazione   (x - 1)(3x + 1) =3D 2x - 1

si riduce in forma normale, ottenendo l’equazione spuria

3x² - 4x =3D 0

si raccoglie x       = ;     x(3x - 4) =3D 0

si applica la legge di annullamento del prodotto ponendo

x =3D 0 Ú 3x - 4 =3D 0

si ricavano due soluzioni reali

x =3D 0 Ú x =3D 4/3

11

Esempio 2

     Ris= olvere l’equazione   = x(x - 1) =3D - x(x + 1) + 8

eseguendo i calcoli e applicando il 1° principio di equivalenza, si ottiene l’equazione pura

2x² - 8 =3D= 0

l’equazione ammette due soluzioni reali e opposte

x² =3D 4

x =3D ± 2

12

Esempio 3

     Ris= olvere l’equazione frazionaria

si applica il 2°principio d’equivalenza, ponendo le condizioni       = ;            &n= bsp; , (*)

si riduce in forma normale, ottenendo l’equazione completa

poiché D > 0 si hann= o due soluzioni reali e distinte

la soluzione x =3D1 non è accettabile per le (*)

13

Risoluzione di un’equazione di 2° grado
Metodo grafico

     L’equazione ax²+ bx + c =3D 0 si ottiene dall’intersezione della parabola di equazione y =3D ax^{2 <=
/sup>+
bx + c con l’asse delle ascisse, la cui equazione è y =
=3D 0.}

     Ese= mpio

     Sia= x² - 4x =3D 0
D > 0

l’equazione ha due soluzioni

     rea= li e distinte x =3D 0 o x =3D 4

la parabola di equazione y =3D x² - 4x ha pertanto due intersezioni con l’a= sse delle ascisse

14

Soluzioni di un’equazione
di 2° grado e scomposizione in fattori

       Data l’equazione ax²+ bx + c =3D 0, si dimo= stra che essa è equivalente a

a(x – x₁)(x – x₂) =3D 0

       dove x₁ ed x₂ sono le soluzioni dell’equazione determinate applicando la formula risolutiva de= lle equazioni complete di 2° grado.

       Esempi

        1. L’equa= zione 4x²- 3x - 1 =3D 0 ha soluzioni x₁=3D -1/4 e<= /li>
       = ;       x₂=3D 1 e pertanto è equivalente a 4(x + ¼) (x– 1) =3D 0.

        2. Viceversa se l’equazione si presenta nella forma

       = ;     3(x+ 2)(x – 5) =3D 0, le soluzioni sono x1 =3D –2 e x_{2=3D 5.}

15

Disequazioni di= 2° grado

    Sono scri= tte nella forma:

ax²+ bx + c < 0 (oppure ax^=
2+ bx + c £ 0)

ax²+ bx + c > 0 (oppure ax^=
2+ bx + c ³ 0)

     Calcolarne le soluzioni significa stabilire per quali valori = di x il trinomio ax²+ bx + c= è negativo, nullo o positivo, cioè per quali va= lori di x la parabola di equazione

y =3D ax²+ bx + c

     &eg= rave; negativa, nulla o positiva.

16

Risoluzione di = una
disequazione di 2° gr= ado

       Se la disequazione è del tipo

ax²+ bx += c > 0 (oppure ax²+ bx + c ³ 0) con a = > 0

        si possono presentare tre casi (D<= font face=3D"Times New Roman"> > 0,&n= bsp; D =3D 0, D < 0).

        Prima di passare all’analisi di questi tre casi si rico= rda che se a > 0 la parabola di equazione y =3D ax²+ bx += c ha la concavità rivolta verso l’alto e che se a < 0 è sempre possibile cambiare il segno a tutti i termini della disequazione (purchè se ne cambi il verso!!) e ricondursi anc= ora al caso a > 0.

17

Risoluzione di = una
disequazione di 2° gr= ado

       ax²+ bx + c > 0 (ax²+ bx + c ³ 0) con a > 0

se D > 0 l’equazione associata ax²+ bx + c =3D 0 ha due sol= uzioni reali e distinte x₁e x₂, quindi la parabola di equazione y =3D ax²+ bx + c interseca l’asse delle ascisse in due punti distinti ed è positiva (positiva o nulla)   se   x < x₁Ú x > x₂(x £ x₁&= Uacute; x ³= x₂)

18

Risoluzione di = una
disequazione di 2° gr= ado

ax²+ bx + c > 0 (ax²+ bx + c ³ 0) con&nb= sp; a > 0

se D =3D 0 l’equazione associata ax²+ bx + c =3D 0 ha due sol= uzioni reali e coincidenti x₁=3D x₂, qui= ndi la parabola di equazione y =3D ax²+ bx + c interseca l̵= 7;asse delle ascisse in un solo punto (è tangente) ed è posit= iva (positiva o nulla) per= ogni valore di x con x ¹ x_{1 <=
/sub>(per ogni valore=
di x)}

19

Risoluzione di = una
disequazione di 2° gr= ado

ax²+ bx + c > 0 (ax²+ bx + c ³ 0) con&nb= sp; a > 0

se D < 0&nb= sp; l’equazione associata&= nbsp; ax²+ bx + c =3D 0 non ha soluzion= i reali,   quindi la parabola di equazione y =3D ax²⁼+ bx + c non interseca l’asse delle ascisse ed è positiva (positiva o nulla) per ogni valore di x (per ogni valore di x)=

20

Esempio 1

     Ris= olvere la disequazione       x²+ 4x + 4 > 0

la parabola di equazione y =3D x²+ 4x + 4 ha la concavità rivolta verso l’alto

l’equazione, associata alla disequazione, x²+ 4x += 4 =3D 0 ha D =3D= 0 e soluzione doppia x =3D= -2

la parabola è allora tangente all’asse x

     nel vertice di ascissa -2

la parabola è pertanto positiva per ogni

     val= ore di x ad esclusione del valore

     x = =3D -2 per cui è nulla

La soluzione è = " x = ¹ -2=

21

Esempio 2

22

Esempio 3

       = ; Risolvere

        Risolvere un sistema di disequazioni significa determinare le soluzioni che esse hanno in comune

       = ;   S1:

       = ;   S2:

       = ;   S3:

       = ;   Soluzione del sistema:

23

Disequazioni di= grado
superiore al secondo

     Per risolvere una disequazione di grado superiore al secondo può essere conveniente, con una fattorizzazione, scriverle nella forma

A(x)×B(x)×C(x) >= 0 (oppure ³)

A(x)×B(x)×C(x) <= 0 (oppure £)

In entrambi i casi, per risolvere:

si pone ogni fattore > 0 (oppure ³ 0)

si risolve ciascuna disequazione e si rappresentano graficamente le soluzioni

la soluzione finale è data dagli intervalli in cui il prodotto dei segni è positivo o negativo (oppure positivo o nullo, negativo o nullo).

24

Esempio

     Ris= olvere la disequazione

si scompone in fattori

si pone ogni fattore &= gt; 0 e si risolve ciascuna disequazione

     1&d= eg; Fattore:

     2&d= eg; Fattore:

     3&d= eg; Fattore:

25

Segue l’e= sempio

si rappresentano graficamente le soluzioni

la soluzione della disequazione è data dagli intervalli in cu= i il prodotto dei segni è positivo

26

Disequazioni frazionarie

     Sono scritte nella forma

       = ;            &n= bsp;             (oppure ³)       = ;              (oppure <= /font>£)<= /li>

     Una disequazione si dice frazionaria se= l’incognita compare al denominatore. In entrambi i casi, per risolvere

si pone il numeratore > 0 (oppure ³ 0)

si pone il denominatore > 0

si risolve ciascuna disequazione e si rappresentano graficamente le soluzioni

la soluzione finale è data dagli intervalli in cui il prodotto dei segni è positivo o negativo (oppure positivo o nullo, negativo o nullo).

27

Esempio

     Ris= olvere la disequazione

    In tal ca= so, poiché il numeratore è scomposto in fattori, si pu&ogr= ave; procedere ponendo ogni fattore > 0

    1° Fa= ttore:

    2° Fa= ttore:

    3° Fa= ttore:

28

Segue l’e= sempio

si rappresentano graficamente le soluzioni

la soluzione della disequazione è data dagli intervalli in cu= i il prodotto dei segni è negativo

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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1	Equazioni – Disequazioni di 2° grad= o
2	Contenuti Forma normale di una equazione di secondo grado Equazioni di secondo grado incomplete - complete Risoluzione di una equazione incompleta - completa Condizioni sul discriminante Risoluzione di una equazione di 2° grado- metodo grafico Soluzioni di un’equazione di 2° grado e scomposizione in fattori Disequazioni di 2° grado Risoluzione di una disequazioni di 2° grado Disequazioni di grado superiore al secondo Disequazioni frazionarie
3	Forma normale Un’equazione di secondo grado, in forma normale, &egrav= e; del tipo = ; &n= bsp; ax²= + bx + c =3D 0 è ordinata in ordine decrescente rispetto le potenze della incognita non presenta termini simili Ese= mpio 4x<= sup>2 - 5x - 1 =3D 0
4	Esempio Scrivere l’equazione : 3x= (2 - x) =3D 2(x + 1) in forma normale si eseguono i calcoli 6x = - 3x² =3D 2x +2 si applica il 1°principio d’equivalenza = ; &n= bsp; = ; 6x – 3x²- 2x - 2 =3D 0 si sommano i termini simili = ; &n= bsp; = ; –3x²+ 4x –2 =3D 0 si applica il 2°principio di equivalenza = ; &n= bsp; = ; 3x²- 4x + 2 =3D 0
5	Equazioni incomp= lete Un’equazione di secondo grado di dice incompleta se man= ca di uno o più termini: se b =3D 0 si h= a ax² + c = =3D 0 e l’equazione si dice pura se c =3D 0 si h= a ax² + bx =3D 0 e l’equazione si dice spuria
6	"ax^{2 + c =3D 0"} ax² + c =3D 0 Applicando i principi di equivalenza si ha se -c/a > 0 allora l’equazione ammette due soluzioni reali e opposte se -c/a =3D 0, cio&egr= ave; c =3D 0, allora l’equazione ammette due soluzioni nulle: in tal caso l’equazione si dice monomia se -c/a < 0 allora l’equazione non ammette soluzioni reali<= /li>
7	"Equazione = spuria" Equazione spuria
8	Equazione comple= ta Per risol= vere un’equazione di 2° grado completa ax²+ bx + = c =3D 0 si utilizza la formula La quantità sotto radice b² – 4ac si chiama discriminante e si indica con la lettera D (leggi “delta”). Tale quantità “discrimina” i vari casi.
9	Condizioni sul discriminante Se D > = 0 si hanno due soluzioni reali e distinte Se D =3D 0= si hanno due soluzioni reali e coincidenti (cioè una soluzione doppia)= Se D < = 0 non si hanno soluzioni reali e l’equazione si dice impossibile in R
10	Esempio 1 Ris= olvere l’equazione (x - 1)(3x + 1) =3D 2x - 1 si riduce in forma normale, ottenendo l’equazione spuria 3x² - 4x =3D 0 si raccoglie x = ; x(3x - 4) =3D 0 si applica la legge di annullamento del prodotto ponendo x =3D 0 Ú 3x - 4 =3D 0 si ricavano due soluzioni reali x =3D 0 Ú x =3D 4/3
11	Esempio 2 Ris= olvere l’equazione = x(x - 1) =3D - x(x + 1) + 8 eseguendo i calcoli e applicando il 1° principio di equivalenza, si ottiene l’equazione pura 2x² - 8 =3D= 0 l’equazione ammette due soluzioni reali e opposte x² =3D 4 x =3D ± 2
12	Esempio 3 Ris= olvere l’equazione frazionaria si applica il 2°principio d’equivalenza, ponendo le condizioni = ; &n= bsp; , () si riduce in forma normale, ottenendo l’equazione completa poiché D > 0 si hann= o due soluzioni reali e distinte la soluzione x =3D1 non è accettabile per le ()
13	Risoluzione di un’equazione di 2° grado Metodo grafico L’equazione ax²+ bx + c =3D 0 si ottiene dall’intersezione della parabola di equazione y =3D ax^{2 <= /sup>+ bx + c con l’asse delle ascisse, la cui equazione è y = =3D 0.} Ese= mpio Sia= x² - 4x =3D 0 D > 0 l’equazione ha due soluzioni rea= li e distinte x =3D 0 o x =3D 4 la parabola di equazione y =3D x² - 4x ha pertanto due intersezioni con l’a= sse delle ascisse
14	Soluzioni di un’equazione di 2° grado e scomposizione in fattori Data l’equazione ax²+ bx + c =3D 0, si dimo= stra che essa è equivalente a a(x – x₁)(x – x₂) =3D 0 dove x₁ ed x₂ sono le soluzioni dell’equazione determinate applicando la formula risolutiva de= lle equazioni complete di 2° grado. Esempi 1. L’equa= zione 4x²- 3x - 1 =3D 0 ha soluzioni x₁=3D -1/4 e<= /li> = ; x₂=3D 1 e pertanto è equivalente a 4(x + ¼) (x– 1) =3D 0. 2. Viceversa se l’equazione si presenta nella forma = ; 3(x+ 2)(x – 5) =3D 0, le soluzioni sono x1 =3D –2 e x_{2=3D 5.}
15	Disequazioni di= 2° grado Sono scri= tte nella forma: ax²+ bx + c < 0 (oppure ax^= 2+ bx + c £ 0) ax²+ bx + c > 0 (oppure ax^= 2+ bx + c ³ 0) Calcolarne le soluzioni significa stabilire per quali valori = di x il trinomio ax²+ bx + c= è negativo, nullo o positivo, cioè per quali va= lori di x la parabola di equazione y =3D ax²+ bx + c &eg= rave; negativa, nulla o positiva.
16	Risoluzione di = una disequazione di 2° gr= ado Se la disequazione è del tipo ax²+ bx += c > 0 (oppure ax²+ bx + c ³ 0) con a = > 0 si possono presentare tre casi (D<= font face=3D"Times New Roman"> > 0,&n= bsp; D =3D 0, D < 0). Prima di passare all’analisi di questi tre casi si rico= rda che se a > 0 la parabola di equazione y =3D ax²+ bx += c ha la concavità rivolta verso l’alto e che se a < 0 è sempre possibile cambiare il segno a tutti i termini della disequazione (purchè se ne cambi il verso!!) e ricondursi anc= ora al caso a > 0.
17	Risoluzione di = una disequazione di 2° gr= ado ax²+ bx + c > 0 (ax²+ bx + c ³ 0) con a > 0 se D > 0 l’equazione associata ax²+ bx + c =3D 0 ha due sol= uzioni reali e distinte x₁e x₂, quindi la parabola di equazione y =3D ax²+ bx + c interseca l’asse delle ascisse in due punti distinti ed è positiva (positiva o nulla) se x < x₁Ú x > x₂(x £ x₁&= Uacute; x ³= x₂)
18	Risoluzione di = una disequazione di 2° gr= ado ax²+ bx + c > 0 (ax²+ bx + c ³ 0) con&nb= sp; a > 0 se D =3D 0 l’equazione associata ax²+ bx + c =3D 0 ha due sol= uzioni reali e coincidenti x₁=3D x₂, qui= ndi la parabola di equazione y =3D ax²+ bx + c interseca l̵= 7;asse delle ascisse in un solo punto (è tangente) ed è posit= iva (positiva o nulla) per= ogni valore di x con x ¹ x_{1 <= /sub>(per ogni valore= di x)}
19	Risoluzione di = una disequazione di 2° gr= ado ax²+ bx + c > 0 (ax²+ bx + c ³ 0) con&nb= sp; a > 0 se D < 0&nb= sp; l’equazione associata&= nbsp; ax²+ bx + c =3D 0 non ha soluzion= i reali, quindi la parabola di equazione y =3D ax²⁼+ bx + c non interseca l’asse delle ascisse ed è positiva (positiva o nulla) per ogni valore di x (per ogni valore di x)=
20	Esempio 1 Ris= olvere la disequazione x²+ 4x + 4 > 0 la parabola di equazione y =3D x²+ 4x + 4 ha la concavità rivolta verso l’alto l’equazione, associata alla disequazione, x²+ 4x += 4 =3D 0 ha D =3D= 0 e soluzione doppia x =3D= -2 la parabola è allora tangente all’asse x nel vertice di ascissa -2 la parabola è pertanto positiva per ogni val= ore di x ad esclusione del valore x = =3D -2 per cui è nulla La soluzione è = " x = ¹ -2=
21	Esempio 2
22	Esempio 3 = ; Risolvere Risolvere un sistema di disequazioni significa determinare le soluzioni che esse hanno in comune = ; S1: = ; S2: = ; S3: = ; Soluzione del sistema:
23	Disequazioni di= grado superiore al secondo Per risolvere una disequazione di grado superiore al secondo può essere conveniente, con una fattorizzazione, scriverle nella forma A(x)×B(x)×C(x) >= 0 (oppure ³) A(x)×B(x)×C(x) <= 0 (oppure £) In entrambi i casi, per risolvere: si pone ogni fattore > 0 (oppure ³ 0) si risolve ciascuna disequazione e si rappresentano graficamente le soluzioni la soluzione finale è data dagli intervalli in cui il prodotto dei segni è positivo o negativo (oppure positivo o nullo, negativo o nullo).
24	Esempio Ris= olvere la disequazione si scompone in fattori si pone ogni fattore &= gt; 0 e si risolve ciascuna disequazione 1&d= eg; Fattore: 2&d= eg; Fattore: 3&d= eg; Fattore:
25	Segue l’e= sempio si rappresentano graficamente le soluzioni la soluzione della disequazione è data dagli intervalli in cu= i il prodotto dei segni è positivo
26	Disequazioni frazionarie Sono scritte nella forma = ; &n= bsp; (oppure ³) = ; (oppure <= /font>£)<= /li> Una disequazione si dice frazionaria se= l’incognita compare al denominatore. In entrambi i casi, per risolvere si pone il numeratore > 0 (oppure ³ 0) si pone il denominatore > 0 si risolve ciascuna disequazione e si rappresentano graficamente le soluzioni la soluzione finale è data dagli intervalli in cui il prodotto dei segni è positivo o negativo (oppure positivo o nullo, negativo o nullo).
27	Esempio Ris= olvere la disequazione In tal ca= so, poiché il numeratore è scomposto in fattori, si pu&ogr= ave; procedere ponendo ogni fattore > 0 1° Fa= ttore: 2° Fa= ttore: 3° Fa= ttore:
28	Segue l’e= sempio si rappresentano graficamente le soluzioni la soluzione della disequazione è data dagli intervalli in cu= i il prodotto dei segni è negativo