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Subject: =?Windows-1252?Q?Risolutore_javascript_di_equazioni_di_2=B0e_3=B0_grado?=
Date: Wed, 3 Sep 2008 16:29:18 +0200
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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD><TITLE>Risolutore javascript di equazioni di 2=B0e 3=B0 =
grado</TITLE>
<META http-equiv=3Dcontent-type content=3D"text/html; =
charset=3Diso-8859-1">
<META content=3D"Gianfranco Bo" name=3DAuthor>
<META content=3DFrontPage.Editor.Document name=3DProgId><LINK=20
href=3D"http://utenti.quipo.it/base5/stileb5.css" type=3Dtext/css =
rel=3Dstylesheet>
<SCRIPT language=3Djavascript type=3Dtext/javascript>=0A=
<!--Define JavaScript functions.-->=0A=
=0A=
=0A=
function Quad2Solve(dataForm){=0A=
var a =3D parseFloat(dataForm.aIn.value);=0A=
var b =3D parseFloat(dataForm.bIn.value);=0A=
var c =3D parseFloat(dataForm.cIn.value);=0A=
=0A=
if (a =3D=3D 0){=0A=
 if (b =3D=3D 0)=0A=
  alert("Dati non accettabili: a, b sono nulli.");=0A=
 else{=0A=
  dataForm.x1Re.value =3D -c/b;=0A=
  dataForm.x1Im.value =3D 0;=0A=
  dataForm.x2Re.value =3D "N/A";=0A=
  dataForm.x2Im.value =3D "N/A";=0A=
 } // End else=0A=
 return;=0A=
} //End if a =3D=3D 0=0A=
=0A=
if (b =3D=3D 0){=0A=
 var dum1 =3D -c/a;=0A=
 if (dum1 < 0){=0A=
  dum1 =3D -dum1;=0A=
  dum1 =3D Math.sqrt(dum1);=0A=
  dataForm.x2Re.value =3D dataForm.x1Re.value =3D 0;=0A=
  dataForm.x1Im.value =3D dataForm.x2Im.value =3D dum1;=0A=
 } //End if dum1 < 0=0A=
 else{=0A=
  dum1 =3D Math.sqrt(dum1);=0A=
  dataForm.x1Re.value =3D dum1;=0A=
  dataForm.x2Re.value =3D -dum1;=0A=
  dataForm.x2Im.value =3D dataForm.x1Im.value =3D 0;=0A=
 } //End else=0A=
 return;=0A=
} //End if b =3D=3D 0=0A=
=0A=
if (!c){=0A=
 dataForm.x2Im.value =3D dataForm.x1Im.value =3D dataForm.x1Re.value =3D =
0;=0A=
 dataForm.x2Re.value =3D -b/a;=0A=
 return;=0A=
}//End c =3D=3D 0=0A=
=0A=
b /=3D 2.0*a;=0A=
c /=3D a;=0A=
=0A=
var discrim;=0A=
=0A=
discrim =3D -c + b*b;=0A=
=0A=
if (discrim < 0){=0A=
 discrim =3D -discrim;=0A=
 dataForm.x2Re.value =3D dataForm.x1Re.value =3D -b;=0A=
 dataForm.x2Im.value =3D dataForm.x1Im.value =3D Math.sqrt(discrim);=0A=
} //End discrim < 0=0A=
else {=0A=
 if (b < 0){=0A=
  b =3D -b;=0A=
  a =3D b + Math.sqrt(discrim);=0A=
 } // End if b < 0=0A=
 else{=0A=
  a =3D b + Math.sqrt(discrim);=0A=
  a =3D -a;=0A=
 } //End else b >=3D 0=0A=
 dataForm.x1Re.value =3D c/a;=0A=
 dataForm.x2Re.value =3D a;=0A=
 dataForm.x2Im.value =3D dataForm.x1Im.value =3D 0;=0A=
} //End else discrim >=3D 0=0A=
return;=0A=
}  //End of Quad2Solve=0A=
=0A=
// end of JavaScript-->=0A=
=0A=
<!--Define JavaScript functions.-->=0A=
=0A=
=0A=
function Quad3Solve(dataForm){=0A=
var a =3D parseFloat(dataForm.aIn.value);=0A=
var b =3D parseFloat(dataForm.bIn.value);=0A=
var c =3D parseFloat(dataForm.cIn.value);=0A=
var d =3D parseFloat(dataForm.dIn.value);=0A=
=0A=
if (a =3D=3D 0){=0A=
 alert("Il coefficiente a =E8 nullo? L'equazione =E8 di 2=B0 grado!");=0A=
 return;=0A=
} //End if a =3D=3D 0=0A=
=0A=
if (d =3D=3D 0){=0A=
 alert("Una radice =E8 0. Dividi per x e utilizza il programma per le =
equazioni di 2=B0 grado.");=0A=
 return;=0A=
} //End if d =3D=3D 0=0A=
=0A=
b /=3D a;=0A=
c /=3D a;=0A=
d /=3D a;=0A=
=0A=
var discrim, q, r, dum1, s, t, term1, r13;=0A=
=0A=
q =3D (3.0*c - (b*b))/9.0;=0A=
r =3D -(27.0*d) + b*(9.0*c - 2.0*(b*b));=0A=
r /=3D 54.0;=0A=
=0A=
discrim =3D q*q*q + r*r;=0A=
dataForm.x1Im.value =3D 0; //The first root is always real.=0A=
term1 =3D (b/3.0);=0A=
=0A=
if (discrim > 0) { // one root real, two are complex=0A=
 s =3D r + Math.sqrt(discrim);=0A=
 s =3D ((s < 0) ? -Math.pow(-s, (1.0/3.0)) : Math.pow(s, (1.0/3.0)));=0A=
 t =3D r - Math.sqrt(discrim);=0A=
 t =3D ((t < 0) ? -Math.pow(-t, (1.0/3.0)) : Math.pow(t, (1.0/3.0)));=0A=
 dataForm.x1Re.value =3D -term1 + s + t;=0A=
 term1 +=3D (s + t)/2.0;=0A=
 dataForm.x3Re.value =3D dataForm.x2Re.value =3D -term1;=0A=
 term1 =3D Math.sqrt(3.0)*(-t + s)/2;=0A=
 dataForm.x2Im.value =3D term1;=0A=
 dataForm.x3Im.value =3D -term1;=0A=
 return;=0A=
} // End if (discrim > 0)=0A=
=0A=
// The remaining options are all real=0A=
dataForm.x3Im.value =3D dataForm.x2Im.value =3D 0;=0A=
=0A=
if (discrim =3D=3D 0){ // All roots real, at least two are equal.=0A=
 r13 =3D ((r < 0) ? -Math.pow(-r,(1.0/3.0)) : Math.pow(r,(1.0/3.0)));=0A=
 dataForm.x1Re.value =3D -term1 + 2.0*r13;=0A=
 dataForm.x3Re.value =3D dataForm.x2Re.value =3D -(r13 + term1);=0A=
 return;=0A=
} // End if (discrim =3D=3D 0)=0A=
=0A=
// Only option left is that all roots are real and unequal (to get here, =
q < 0)=0A=
q =3D -q;=0A=
dum1 =3D q*q*q;=0A=
dum1 =3D Math.acos(r/Math.sqrt(dum1));=0A=
r13 =3D 2.0*Math.sqrt(q);=0A=
dataForm.x1Re.value =3D -term1 + r13*Math.cos(dum1/3.0);=0A=
dataForm.x2Re.value =3D -term1 + r13*Math.cos((dum1 + 2.0*Math.PI)/3.0);=0A=
dataForm.x3Re.value =3D -term1 + r13*Math.cos((dum1 + 4.0*Math.PI)/3.0);=0A=
return;=0A=
=0A=
}  //End of Quad3Solve=0A=
// end of JavaScript-->=0A=
=0A=
<!--Define JavaScript functions.-->=0A=
=0A=
=0A=
function Quad4Solve(dataForm){=0A=
var a =3D parseFloat(dataForm.aIn.value);=0A=
var b =3D parseFloat(dataForm.bIn.value);=0A=
var c =3D parseFloat(dataForm.cIn.value);=0A=
var d =3D parseFloat(dataForm.dIn.value);=0A=
var e =3D parseFloat(dataForm.eIn.value);=0A=
=0A=
if (a =3D=3D 0){=0A=
 alert("Il coefficiente a =E8 nullo? L'equazione =E8 di terzo grado!");=0A=
 return;=0A=
} //End if a =3D=3D 0=0A=
=0A=
if (e =3D=3D 0){=0A=
 alert("Una radice =E8 0. Dividi per x e utilizza il programma per le =
equazioni di 3=B0 grado.");=0A=
 return;=0A=
} //End if e =3D=3D 0=0A=
=0A=
if (a !=3D 1) {=0A=
 b /=3D a;=0A=
 c /=3D a;=0A=
 d /=3D a;=0A=
 e /=3D a;=0A=
}=0A=
=0A=
var cb, cc, cd;  // Coefficients for use with cubic solver=0A=
var discrim, q, r, RRe, RIm, DRe, DIm, dum1, ERe, EIm, s, t, term1, r13, =
sqR, y1, z1Re, z1Im, z2Re;=0A=
=0A=
cb =3D -c;=0A=
cc =3D -4.0*e + d*b;=0A=
cd =3D -(b*b*e + d*d) + 4.0*c*e;=0A=
=0A=
if (cd =3D=3D 0)  alert("cd =3D 0.");=0A=
=0A=
// Solve the resolvant cubic for y1=0A=
q =3D (3.0*cc - (cb*cb))/9.0;=0A=
r =3D -(27.0*cd) + cb*(9.0*cc - 2.0*(cb*cb));=0A=
r /=3D 54.0;=0A=
discrim =3D q*q*q + r*r;=0A=
term1 =3D (cb/3.0);=0A=
=0A=
if (discrim > 0) { // one root real, two are complex=0A=
 s =3D r + Math.sqrt(discrim);=0A=
 s =3D ((s < 0) ? -Math.pow(-s, (1.0/3.0)) : Math.pow(s, (1.0/3.0)));=0A=
 t =3D r - Math.sqrt(discrim);=0A=
 t =3D ((t < 0) ? -Math.pow(-t, (1.0/3.0)) : Math.pow(t, (1.0/3.0)));=0A=
 y1 =3D -term1 + s + t;=0A=
} // End if (discrim > 0)=0A=
else {=0A=
 if (discrim =3D=3D 0) {=0A=
  r13 =3D ((r < 0) ? -Math.pow(-r,(1.0/3.0)) : Math.pow(r,(1.0/3.0)));=0A=
  y1 =3D -term1 + 2.0*r13;=0A=
 } // End if (discrim =3D=3D 0)=0A=
 else { // else discrim < 0=0A=
  q =3D -q;=0A=
  dum1 =3D q*q*q;=0A=
  dum1 =3D Math.acos(r/Math.sqrt(dum1));=0A=
  r13 =3D 2.0*Math.sqrt(q);=0A=
  y1 =3D -term1 + r13*Math.cos(dum1/3.0);=0A=
 } // End discrim < 0=0A=
} // End else discrim <=3D 0=0A=
=0A=
// At this point, we have determined y1, a real root of the resolvent =
cubic.=0A=
// Carry on to solve the original quartic equation=0A=
=0A=
term1 =3D b/4.0;=0A=
sqR =3D -c + term1*b + y1;  // R-squared=0A=
=0A=
RRe =3D RIm =3D DRe =3D DIm =3D ERe =3D EIm =3D z1Re =3D z1Im =3D z2Re =
=3D 0;=0A=
=0A=
if (sqR >=3D 0) {=0A=
 if (sqR =3D=3D 0) {=0A=
  dum1 =3D -(4.0*e) + y1*y1;=0A=
  if (dum1 < 0) //D and E will be complex=0A=
   z1Im =3D 2.0*Math.sqrt(-dum1);=0A=
  else { //else (dum1 >=3D 0)=0A=
   z1Re =3D 2.0*Math.sqrt(dum1);=0A=
   z2Re =3D -z1Re;=0A=
  }//End else (dum1 >=3D 0)=0A=
 } //End if (sqR =3D=3D 0)=0A=
 else { //(sqR > 0)=0A=
  RRe =3D Math.sqrt(sqR);=0A=
  z1Re =3D -(8.0*d + b*b*b)/4.0 + b*c;=0A=
  z1Re /=3D RRe;=0A=
  z2Re =3D -z1Re;=0A=
 } // End else (sqR > 0)=0A=
} //end if (sqR >=3D 0)=0A=
else { //else (sqR < 0)=0A=
 RIm =3D Math.sqrt(-sqR);=0A=
 z1Im =3D -(8.0*d + b*b*b)/4.0 + b*c;=0A=
 z1Im /=3D RIm;=0A=
 z1Im =3D -z1Im;=0A=
} // End else (sqR < 0)=0A=
=0A=
z1Re +=3D -(2.0*c + sqR) + 3.0*b*term1;=0A=
z2Re +=3D -(2.0*c + sqR) + 3.0*b*term1;=0A=
=0A=
//At this point, z1 and z2 should be the terms under the square root for =
D and E=0A=
=0A=
if (z1Im =3D=3D 0){ // Both z1 and z2 real=0A=
 if (z1Re >=3D 0)=0A=
  DRe =3D Math.sqrt(z1Re);=0A=
 else=0A=
  DIm =3D Math.sqrt(-z1Re);=0A=
 if (z2Re >=3D 0)=0A=
  ERe =3D Math.sqrt(z2Re);=0A=
 else=0A=
  EIm =3D Math.sqrt(-z2Re);=0A=
}// End if (zIm =3D=3D 0)=0A=
else { //else (zIm !=3D 0); calculate root of a complex number********=0A=
 r =3D Math.sqrt(z1Re*z1Re + z1Im*z1Im); // Calculate r, the magnitude=0A=
 r =3D Math.sqrt(r);=0A=
=0A=
 dum1 =3D Math.atan2(z1Im, z1Re); // Calculate the angle between the two =
vectors=0A=
 dum1 /=3D 2; //Divide this angle by 2=0A=
 ERe =3D DRe =3D r*Math.cos(dum1); //Form the new complex value=0A=
 DIm =3D r*Math.sin(dum1);=0A=
 EIm =3D -DIm;=0A=
} // End else (z1Im !=3D 0)=0A=
=0A=
dataForm.x1Re.value =3D -term1 + (RRe + DRe)/2;=0A=
dataForm.x1Im.value =3D (RIm + DIm)/2;=0A=
dataForm.x2Re.value =3D -(term1 + DRe/2) + RRe/2;=0A=
dataForm.x2Im.value =3D (-DIm + RIm)/2;=0A=
dataForm.x3Re.value =3D -(term1 + RRe/2) + ERe/2;=0A=
dataForm.x3Im.value =3D (-RIm + EIm)/2;=0A=
dataForm.x4Re.value =3D -(term1 + (RRe + ERe)/2);=0A=
dataForm.x4Im.value =3D -(RIm + EIm)/2;=0A=
=0A=
return;=0A=
}  //End of Quad4Solve=0A=
// end of JavaScript-->=0A=
=0A=
  </SCRIPT>

<META content=3D"MSHTML 6.00.6000.16705" name=3DGENERATOR></HEAD>
<BODY>
<P align=3Dleft>[<A =
href=3D"http://utenti.quipo.it/base5/index.htm"><B>HOME - BASE=20
Cinque</B> - <I>Appunti di Matematica ricreativa</I></A>]</P>
<H1 align=3Dcenter>Risolutore di equazioni di 2=B0 e 3=B0 grado</H1>
<P align=3Dcenter><B>Un programma Javascript per risolvere equazioni =
numeriche di=20
2=B0 e 3=B0 grado</B></P>
<P align=3Dleft>Questa pagina contiene codice javascript. Potete =
scaricarla sul=20
vostro computer e utilizzarla senza essere collegati ad internet.</P>
<HR color=3D#ff0000>

<TABLE cellSpacing=3D0 cellPadding=3D3 width=3D"100%" bgColor=3D#ffffcc =
border=3D1>
  <TBODY>
  <TR>
    <TD width=3D"100%"><B><FONT color=3D#ff0000 size=3D4>Equazioni di =
2=B0=20
      grado</FONT></B>=20
      <P>Un'equazione di 2=B0 grado =E8 della forma:<BR><B>a</B> x =
<SUP>2</SUP> +=20
      <B>b</B> x + <B>c</B> =3D 0<BR>dove a, b, c, sono i coefficienti e =
x =E8=20
      l'incognita.<BR>Le equazioni di secondo grado hanno due=20
      soluzioni.<BR>Inserite i valori numerici dei coefficienti a, b, c, =
nel=20
      quadro qui sotto e cliccate su "Risolvi!"<BR><B>Nota: se un =
coefficiente =E8=20
      nullo, bisogna inserire il numero "0"</B></P>
      <FORM name=3Dform1>
      <P><B>a</B>: <INPUT size=3D8 name=3DaIn> <B>b</B>: <INPUT size=3D8 =
name=3DbIn>=20
      <B>c</B>: <INPUT size=3D8 name=3DcIn></P>
      <P><INPUT onclick=3DQuad2Solve(this.form) type=3Dbutton =
value=3DRisolvi! name=3Dbutton1></P>
      <P>Le soluzioni sono:<BR><FONT size=3D2>(Nota: sono indicate la =
parte reale=20
      e quella immaginaria. Le radici reali sono quelle che hanno la =
parte=20
      immaginaria uguale a 0)</FONT></P>
      <P>x<SUB>1</SUB>: <INPUT name=3Dx1Re> + &nbsp;<INPUT name=3Dx1Im>=20
      i<BR>x<SUB>2</SUB>: <INPUT name=3Dx2Re> - &nbsp;<INPUT =
name=3Dx2Im>=20
      i</P></FORM></TD></TR></TBODY></TABLE>
<HR color=3D#ff0000>

<TABLE cellSpacing=3D0 cellPadding=3D3 width=3D"100%" bgColor=3D#ccffcc =
border=3D1>
  <TBODY>
  <TR>
    <TD width=3D"100%"><B><FONT color=3D#ff0000 size=3D4>Equazioni di =
3=B0=20
      grado</FONT></B>=20
      <P>Un'equazione di 3=B0 grado =E8 della forma:<BR><B>a</B> x =
<SUP>3</SUP> +=20
      <B>b</B> x <SUP>2</SUP> + <B>c</B> x + &nbsp;<B>d</B> =3D =
0<BR>dove a, b, c,=20
      d, sono i coefficienti e x =E8 l'incognita.<BR>Le equazioni di =
terzo grado=20
      hanno tre soluzioni.<BR>Inserite i valori numerici dei =
coefficienti a, b,=20
      c, d, nel quadro qui sotto e cliccate su "Risolvi!"<BR><B>Nota: se =
un=20
      coefficiente =E8 nullo, bisogna inserire il numero "0"</B></P>
      <FORM name=3Dform2><B>a</B>: <INPUT size=3D8 name=3DaIn> <B>b</B>: =
<INPUT size=3D8=20
      name=3DbIn> <B>c</B>: <INPUT size=3D8 name=3DcIn> <B>d</B>: <INPUT =
size=3D8=20
      name=3DdIn>=20
      <P><INPUT onclick=3DQuad3Solve(this.form) type=3Dbutton =
value=3DRisolvi! name=3Dbutton1></P>
      <P>Le soluzioni sono:<BR><FONT size=3D2>(Nota: sono indicate la =
parte reale=20
      e quella immaginaria. Le radici reali sono quelle che hanno la =
parte=20
      immaginaria uguale a 0)</FONT></P>
      <P>x<SUB>1</SUB>: <INPUT name=3Dx1Re> + &nbsp;<INPUT name=3Dx1Im>=20
      i<BR>x<SUB>2</SUB>: <INPUT name=3Dx2Re> + &nbsp;<INPUT =
name=3Dx2Im>=20
      i<BR>x<SUB>3</SUB>: <INPUT name=3Dx3Re> + &nbsp;<INPUT =
name=3Dx3Im>=20
      i</P></FORM></TD></TR></TBODY></TABLE>
<HR color=3D#ff0000>

<P><B>E le equazioni di 4=B0 grado?<BR></B>Il programma che si trovava =
in questa=20
pagina non era molto affidabile, come mi ha segnalato <B>Attilio =
Scifoni</B>,=20
che ringrazio di cuore.<BR>Ho pertanto creato una nuova pagina dedicata=20
esclusivamente a questo argomento.</P>
<P>Se vi interessa, andate a: <A=20
href=3D"http://utenti.quipo.it/base5/numeri/equasolutore4.html">Risolutor=
e di=20
equazioni di 4=B0 grado</A>. </P>
<HR>

<H1 style=3D"TEXT-ALIGN: center">Risposte &amp; riflessioni</H1>
<P align=3Dleft>A titolo d'informazione, riporto le formule per =
risolvere le=20
equazioni di 2=B0, 3=B0 e 4=B0 grado.<BR>Dal 5=B0 grado in avanti non =
esistono formule=20
generali per radicali.</P>
<P align=3Dleft>Equazione di 2=B0 grado: <B>a</B> x <SUP>2</SUP> + =
<B>b</B> x +=20
<B>c</B> =3D 0</P>
<P align=3Dcenter><IMG height=3D61=20
src=3D"http://utenti.quipo.it/base5/numeri/equsol1.gif" width=3D206 =
border=3D0></P>
<P align=3Dleft>Equazione di 3=B0 grado: <B>a</B> x <SUP>3</SUP> + =
<B>b</B> x=20
<SUP>2</SUP> + <B>c</B> x + &nbsp;<B>d</B> =3D 0</P>
<P align=3Dcenter><IMG height=3D176=20
src=3D"http://utenti.quipo.it/base5/numeri/equsol2.gif" width=3D645 =
border=3D0></P>
<P align=3Dleft><I>Data creazione: marzo 2005</I></P>
<P align=3Dleft><I>Ultimo aggiornamento: novembre 2007</I></P>
<HR>

<P style=3D"TEXT-ALIGN: center">Sito Web realizzato da <B>Gianfranco=20
Bo</B></P></BODY></HTML>

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