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How to write Quadratic equation with negative coefficient



The Next CEO of Stack Overflowdef function with arithmetic macroHow to seed FPseed with a value that is unique for every compilation?Select largest value from a list of variables in LaTeXError in pgfplot when big offset is subtractedoverloading functions of the fp packageProblem with negative “multido” and “fp”










11















How to write Quadratic equation with negative coefficient in fp





For a=1, b=-5, c= 6, $ca x^2+cb x + cc=0$ gives 1x^2 + -5x + 6



But i want to have x^2 -5x + 6




documentclassbeamer
usepackagefp
begindocument
beginframeQuadratic equation
FPsetca1
FPsetcb-5
FPsetcc6
FPqsolvexonextwocacbcc
FPevalxoneclip(round(xone:4))
FPevalxtwoclip(round(xtwo:4))
Quadratic equation : $ca x^2+cb x + cc=0$\[1cm]
Result: $x = xone quad textand quad x = xtwo$
endframe
enddocument









share|improve this question




























    11















    How to write Quadratic equation with negative coefficient in fp





    For a=1, b=-5, c= 6, $ca x^2+cb x + cc=0$ gives 1x^2 + -5x + 6



    But i want to have x^2 -5x + 6




    documentclassbeamer
    usepackagefp
    begindocument
    beginframeQuadratic equation
    FPsetca1
    FPsetcb-5
    FPsetcc6
    FPqsolvexonextwocacbcc
    FPevalxoneclip(round(xone:4))
    FPevalxtwoclip(round(xtwo:4))
    Quadratic equation : $ca x^2+cb x + cc=0$\[1cm]
    Result: $x = xone quad textand quad x = xtwo$
    endframe
    enddocument









    share|improve this question


























      11












      11








      11


      2






      How to write Quadratic equation with negative coefficient in fp





      For a=1, b=-5, c= 6, $ca x^2+cb x + cc=0$ gives 1x^2 + -5x + 6



      But i want to have x^2 -5x + 6




      documentclassbeamer
      usepackagefp
      begindocument
      beginframeQuadratic equation
      FPsetca1
      FPsetcb-5
      FPsetcc6
      FPqsolvexonextwocacbcc
      FPevalxoneclip(round(xone:4))
      FPevalxtwoclip(round(xtwo:4))
      Quadratic equation : $ca x^2+cb x + cc=0$\[1cm]
      Result: $x = xone quad textand quad x = xtwo$
      endframe
      enddocument









      share|improve this question
















      How to write Quadratic equation with negative coefficient in fp





      For a=1, b=-5, c= 6, $ca x^2+cb x + cc=0$ gives 1x^2 + -5x + 6



      But i want to have x^2 -5x + 6




      documentclassbeamer
      usepackagefp
      begindocument
      beginframeQuadratic equation
      FPsetca1
      FPsetcb-5
      FPsetcc6
      FPqsolvexonextwocacbcc
      FPevalxoneclip(round(xone:4))
      FPevalxtwoclip(round(xtwo:4))
      Quadratic equation : $ca x^2+cb x + cc=0$\[1cm]
      Result: $x = xone quad textand quad x = xtwo$
      endframe
      enddocument






      fp






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Mar 21 at 3:52







      sandu

















      asked Mar 20 at 9:54









      sandusandu

      3,72342856




      3,72342856




















          3 Answers
          3






          active

          oldest

          votes


















          10














          Some comparison are necessary. This assumes the coefficients are integers.



          documentclassbeamer
          usepackagefp

          newcommandquadratic[4][x]%
          FPsetca#2%
          FPsetcb#3%
          FPsetcc#4%
          FPqsolvexonextwocacbcc
          FPevalxoneclip(round(xone:4))%
          FPevalxtwoclip(round(xtwo:4))%
          Quadratic equation: $
          ifnumca=1
          else
          ifnumca=-1
          -%
          else
          ca
          fi
          fi
          #1^2%
          ifnumcb=0
          else
          ifnumcb>0
          +%
          ifnumcb=1
          else
          cb
          fi
          else
          ifnumcb=-1
          -%
          else
          cb
          fi
          fi
          #1%
          fi
          ifnumcc=0
          else
          ifnumcc>0
          +
          fi
          cc
          fi
          $\[bigskipamount]
          Result: $#1=xone$ and $#1=xtwo$%


          begindocument
          beginframeQuadratic equation

          quadratic1-56

          bigskip

          quadratic[t]231

          bigskip

          quadratic20-8

          endframe
          enddocument


          With expl3:



          documentclassbeamer
          usepackagexparse

          ExplSyntaxOn

          NewDocumentCommandquadraticOxmmm

          Quadratic~equation:~$
          str_case:nnF #2

          1
          -1-

          #2
          #1^2
          str_case:nnF #3

          0
          1+#1
          -1-#1

          fp_compare:nT #3>0 + #3#1
          fp_compare:nF #4 = 0

          fp_compare:nT #4 > 0 +

          #4
          $\[bigskipamount]
          Result:~$#1=sandu_solve:nnnn+#2#3#4$~and~
          $#1=sandu_solve:nnnn-#2#3#4$

          cs_new:Nn sandu_solve:nnnn

          fp_eval:n round( ( -(#3) #1 sqrt((#3)^2-4*(#2)*(#4)) )/(2*(#2)), 4)

          ExplSyntaxOff

          begindocument
          beginframeQuadratic equation

          quadratic1-56

          bigskip

          quadratic[t]231

          bigskip

          quadratic20-8

          endframe
          enddocument


          expl3






          share|improve this answer
































            6














            Will also work with addterm -5x in addition to the intended addtermcb x.



            The addterm macro takes a single argument, expands it once, and passes it to addtermaux. The addtermaux definition will grab the first token of the argument and examine to see if it is a minus - character. If so, it typesets a - and the rest of the argument. If not, it sees whether the first token was a + character. If so, it typesets a + and the rest of the argument. If neither of the above cases apply, it typesets a + and the complete argument.



            In this way, the right output is provided whether cc is set to 6 or set to +6.



            documentclassbeamer
            usepackagefp
            newcommandaddterm[1]expandafteraddtermaux#1relax
            defaddtermaux#1#2relaxifx-#1-#2elseifx+#1+#2else+#1#2fifi
            begindocument
            beginframeQuadratic equation
            FPsetca1
            FPsetcb-5
            FPsetcc6
            FPqsolvexonextwocacbcc
            FPevalxoneclip(round(xone:4))
            FPevalxtwoclip(round(xtwo:4))
            Quadratic equation : $ca x^2 addtermcb x addtermcc=0$\[1cm]
            Result: $x = xone quad textand quad x = xtwo$
            endframe
            enddocument


            enter image description here






            share|improve this answer

























            • could you explain newcommand and def...

              – sandu
              Mar 20 at 11:27











            • @sandu I have edited the answer to provide context.

              – Steven B. Segletes
              Mar 20 at 11:32


















            4















            Edit: See below an improved version.




            Note the [fragile] in beginframe. Necessary with FPifpos.



            documentclassbeamer
            usepackagefp
            begindocument
            beginframe[fragile]Quadratic equation
            FPsetca1
            FPsetcb-5
            FPsetcc6
            FPqsolvexonextwocacbcc
            FPevalxoneclip(round(xone:4))
            FPevalxtwoclip(round(xtwo:4))
            FPevalbabsclip(round(abs(cb):4))
            FPevalcabsclip(round(abs(cc):4))

            Quadratic equation : $ca x^2$ FPifposcb $+$ else $-$ fi $babs x$ FPifposcc $+$ else $-$ fi $cabs=0$ %\[1cm]

            Result: $x = xone quad textand quad x = xtwo$


            endframe
            enddocument


            enter image description here



            Improved version



            This version handle better special situations (when some coefficients of the equation became -1, 1 or 0).



            As fp's FPqsolve doesn't handle equations without solutions (it emit an error), my code don't display correctly equations where the "x^2" AND the "x" term are null (it display, when the compiler don't stop at errors, something like: Quadratic equation: +6 = 0). This code is intended to be used only when the equation has real(s) solution(s).



            documentclassbeamer
            usepackagefp
            begindocument
            beginframe[fragile]Quadratic equation
            FPsetca1
            FPsetcb-5
            FPsetcc6
            FPqsolvexonextwocacbcc
            FPevalxoneclip(round(xone:4))
            FPevalxtwoclip(round(xtwo:4))
            FPevalaabsclip(round(abs(ca):4))
            FPevalbabsclip(round(abs(cb):4))
            FPevalcabsclip(round(abs(cc):4))
            newcommandsignaFPifnegca -elsefi
            newcommandpositiveSignBWithAFPifzeroca else +fi % if ca is 0, no positive sign before the "x" term if cb is positive
            newcommandsignbFPifnegcb -else positiveSignBWithAfi
            newcommandsigncFPifnegcc -else +fi
            newcommandcoeffaFPifeqaabs1 elseaabsfi
            newcommandcoeffbFPifeqbabs1 elsebabsfi
            newcommandpolyaFPifzeroca elsesignacoeffa x^2fi
            newcommandpolybFPifzerocb elsesignbcoeffb xfi
            newcommandpolycFPifzerocc elsesignccabsfi

            Quadratic equation : $polya polyb polyc =0$ \[1cm]

            Result: $x = xone quad textand quad x = xtwo$

            endframe
            enddocument


            enter image description here






            share|improve this answer

























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              3 Answers
              3






              active

              oldest

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              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              10














              Some comparison are necessary. This assumes the coefficients are integers.



              documentclassbeamer
              usepackagefp

              newcommandquadratic[4][x]%
              FPsetca#2%
              FPsetcb#3%
              FPsetcc#4%
              FPqsolvexonextwocacbcc
              FPevalxoneclip(round(xone:4))%
              FPevalxtwoclip(round(xtwo:4))%
              Quadratic equation: $
              ifnumca=1
              else
              ifnumca=-1
              -%
              else
              ca
              fi
              fi
              #1^2%
              ifnumcb=0
              else
              ifnumcb>0
              +%
              ifnumcb=1
              else
              cb
              fi
              else
              ifnumcb=-1
              -%
              else
              cb
              fi
              fi
              #1%
              fi
              ifnumcc=0
              else
              ifnumcc>0
              +
              fi
              cc
              fi
              $\[bigskipamount]
              Result: $#1=xone$ and $#1=xtwo$%


              begindocument
              beginframeQuadratic equation

              quadratic1-56

              bigskip

              quadratic[t]231

              bigskip

              quadratic20-8

              endframe
              enddocument


              With expl3:



              documentclassbeamer
              usepackagexparse

              ExplSyntaxOn

              NewDocumentCommandquadraticOxmmm

              Quadratic~equation:~$
              str_case:nnF #2

              1
              -1-

              #2
              #1^2
              str_case:nnF #3

              0
              1+#1
              -1-#1

              fp_compare:nT #3>0 + #3#1
              fp_compare:nF #4 = 0

              fp_compare:nT #4 > 0 +

              #4
              $\[bigskipamount]
              Result:~$#1=sandu_solve:nnnn+#2#3#4$~and~
              $#1=sandu_solve:nnnn-#2#3#4$

              cs_new:Nn sandu_solve:nnnn

              fp_eval:n round( ( -(#3) #1 sqrt((#3)^2-4*(#2)*(#4)) )/(2*(#2)), 4)

              ExplSyntaxOff

              begindocument
              beginframeQuadratic equation

              quadratic1-56

              bigskip

              quadratic[t]231

              bigskip

              quadratic20-8

              endframe
              enddocument


              expl3






              share|improve this answer





























                10














                Some comparison are necessary. This assumes the coefficients are integers.



                documentclassbeamer
                usepackagefp

                newcommandquadratic[4][x]%
                FPsetca#2%
                FPsetcb#3%
                FPsetcc#4%
                FPqsolvexonextwocacbcc
                FPevalxoneclip(round(xone:4))%
                FPevalxtwoclip(round(xtwo:4))%
                Quadratic equation: $
                ifnumca=1
                else
                ifnumca=-1
                -%
                else
                ca
                fi
                fi
                #1^2%
                ifnumcb=0
                else
                ifnumcb>0
                +%
                ifnumcb=1
                else
                cb
                fi
                else
                ifnumcb=-1
                -%
                else
                cb
                fi
                fi
                #1%
                fi
                ifnumcc=0
                else
                ifnumcc>0
                +
                fi
                cc
                fi
                $\[bigskipamount]
                Result: $#1=xone$ and $#1=xtwo$%


                begindocument
                beginframeQuadratic equation

                quadratic1-56

                bigskip

                quadratic[t]231

                bigskip

                quadratic20-8

                endframe
                enddocument


                With expl3:



                documentclassbeamer
                usepackagexparse

                ExplSyntaxOn

                NewDocumentCommandquadraticOxmmm

                Quadratic~equation:~$
                str_case:nnF #2

                1
                -1-

                #2
                #1^2
                str_case:nnF #3

                0
                1+#1
                -1-#1

                fp_compare:nT #3>0 + #3#1
                fp_compare:nF #4 = 0

                fp_compare:nT #4 > 0 +

                #4
                $\[bigskipamount]
                Result:~$#1=sandu_solve:nnnn+#2#3#4$~and~
                $#1=sandu_solve:nnnn-#2#3#4$

                cs_new:Nn sandu_solve:nnnn

                fp_eval:n round( ( -(#3) #1 sqrt((#3)^2-4*(#2)*(#4)) )/(2*(#2)), 4)

                ExplSyntaxOff

                begindocument
                beginframeQuadratic equation

                quadratic1-56

                bigskip

                quadratic[t]231

                bigskip

                quadratic20-8

                endframe
                enddocument


                expl3






                share|improve this answer



























                  10












                  10








                  10







                  Some comparison are necessary. This assumes the coefficients are integers.



                  documentclassbeamer
                  usepackagefp

                  newcommandquadratic[4][x]%
                  FPsetca#2%
                  FPsetcb#3%
                  FPsetcc#4%
                  FPqsolvexonextwocacbcc
                  FPevalxoneclip(round(xone:4))%
                  FPevalxtwoclip(round(xtwo:4))%
                  Quadratic equation: $
                  ifnumca=1
                  else
                  ifnumca=-1
                  -%
                  else
                  ca
                  fi
                  fi
                  #1^2%
                  ifnumcb=0
                  else
                  ifnumcb>0
                  +%
                  ifnumcb=1
                  else
                  cb
                  fi
                  else
                  ifnumcb=-1
                  -%
                  else
                  cb
                  fi
                  fi
                  #1%
                  fi
                  ifnumcc=0
                  else
                  ifnumcc>0
                  +
                  fi
                  cc
                  fi
                  $\[bigskipamount]
                  Result: $#1=xone$ and $#1=xtwo$%


                  begindocument
                  beginframeQuadratic equation

                  quadratic1-56

                  bigskip

                  quadratic[t]231

                  bigskip

                  quadratic20-8

                  endframe
                  enddocument


                  With expl3:



                  documentclassbeamer
                  usepackagexparse

                  ExplSyntaxOn

                  NewDocumentCommandquadraticOxmmm

                  Quadratic~equation:~$
                  str_case:nnF #2

                  1
                  -1-

                  #2
                  #1^2
                  str_case:nnF #3

                  0
                  1+#1
                  -1-#1

                  fp_compare:nT #3>0 + #3#1
                  fp_compare:nF #4 = 0

                  fp_compare:nT #4 > 0 +

                  #4
                  $\[bigskipamount]
                  Result:~$#1=sandu_solve:nnnn+#2#3#4$~and~
                  $#1=sandu_solve:nnnn-#2#3#4$

                  cs_new:Nn sandu_solve:nnnn

                  fp_eval:n round( ( -(#3) #1 sqrt((#3)^2-4*(#2)*(#4)) )/(2*(#2)), 4)

                  ExplSyntaxOff

                  begindocument
                  beginframeQuadratic equation

                  quadratic1-56

                  bigskip

                  quadratic[t]231

                  bigskip

                  quadratic20-8

                  endframe
                  enddocument


                  expl3






                  share|improve this answer















                  Some comparison are necessary. This assumes the coefficients are integers.



                  documentclassbeamer
                  usepackagefp

                  newcommandquadratic[4][x]%
                  FPsetca#2%
                  FPsetcb#3%
                  FPsetcc#4%
                  FPqsolvexonextwocacbcc
                  FPevalxoneclip(round(xone:4))%
                  FPevalxtwoclip(round(xtwo:4))%
                  Quadratic equation: $
                  ifnumca=1
                  else
                  ifnumca=-1
                  -%
                  else
                  ca
                  fi
                  fi
                  #1^2%
                  ifnumcb=0
                  else
                  ifnumcb>0
                  +%
                  ifnumcb=1
                  else
                  cb
                  fi
                  else
                  ifnumcb=-1
                  -%
                  else
                  cb
                  fi
                  fi
                  #1%
                  fi
                  ifnumcc=0
                  else
                  ifnumcc>0
                  +
                  fi
                  cc
                  fi
                  $\[bigskipamount]
                  Result: $#1=xone$ and $#1=xtwo$%


                  begindocument
                  beginframeQuadratic equation

                  quadratic1-56

                  bigskip

                  quadratic[t]231

                  bigskip

                  quadratic20-8

                  endframe
                  enddocument


                  With expl3:



                  documentclassbeamer
                  usepackagexparse

                  ExplSyntaxOn

                  NewDocumentCommandquadraticOxmmm

                  Quadratic~equation:~$
                  str_case:nnF #2

                  1
                  -1-

                  #2
                  #1^2
                  str_case:nnF #3

                  0
                  1+#1
                  -1-#1

                  fp_compare:nT #3>0 + #3#1
                  fp_compare:nF #4 = 0

                  fp_compare:nT #4 > 0 +

                  #4
                  $\[bigskipamount]
                  Result:~$#1=sandu_solve:nnnn+#2#3#4$~and~
                  $#1=sandu_solve:nnnn-#2#3#4$

                  cs_new:Nn sandu_solve:nnnn

                  fp_eval:n round( ( -(#3) #1 sqrt((#3)^2-4*(#2)*(#4)) )/(2*(#2)), 4)

                  ExplSyntaxOff

                  begindocument
                  beginframeQuadratic equation

                  quadratic1-56

                  bigskip

                  quadratic[t]231

                  bigskip

                  quadratic20-8

                  endframe
                  enddocument


                  expl3







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Mar 20 at 11:44

























                  answered Mar 20 at 11:27









                  egregegreg

                  730k8819283242




                  730k8819283242





















                      6














                      Will also work with addterm -5x in addition to the intended addtermcb x.



                      The addterm macro takes a single argument, expands it once, and passes it to addtermaux. The addtermaux definition will grab the first token of the argument and examine to see if it is a minus - character. If so, it typesets a - and the rest of the argument. If not, it sees whether the first token was a + character. If so, it typesets a + and the rest of the argument. If neither of the above cases apply, it typesets a + and the complete argument.



                      In this way, the right output is provided whether cc is set to 6 or set to +6.



                      documentclassbeamer
                      usepackagefp
                      newcommandaddterm[1]expandafteraddtermaux#1relax
                      defaddtermaux#1#2relaxifx-#1-#2elseifx+#1+#2else+#1#2fifi
                      begindocument
                      beginframeQuadratic equation
                      FPsetca1
                      FPsetcb-5
                      FPsetcc6
                      FPqsolvexonextwocacbcc
                      FPevalxoneclip(round(xone:4))
                      FPevalxtwoclip(round(xtwo:4))
                      Quadratic equation : $ca x^2 addtermcb x addtermcc=0$\[1cm]
                      Result: $x = xone quad textand quad x = xtwo$
                      endframe
                      enddocument


                      enter image description here






                      share|improve this answer

























                      • could you explain newcommand and def...

                        – sandu
                        Mar 20 at 11:27











                      • @sandu I have edited the answer to provide context.

                        – Steven B. Segletes
                        Mar 20 at 11:32















                      6














                      Will also work with addterm -5x in addition to the intended addtermcb x.



                      The addterm macro takes a single argument, expands it once, and passes it to addtermaux. The addtermaux definition will grab the first token of the argument and examine to see if it is a minus - character. If so, it typesets a - and the rest of the argument. If not, it sees whether the first token was a + character. If so, it typesets a + and the rest of the argument. If neither of the above cases apply, it typesets a + and the complete argument.



                      In this way, the right output is provided whether cc is set to 6 or set to +6.



                      documentclassbeamer
                      usepackagefp
                      newcommandaddterm[1]expandafteraddtermaux#1relax
                      defaddtermaux#1#2relaxifx-#1-#2elseifx+#1+#2else+#1#2fifi
                      begindocument
                      beginframeQuadratic equation
                      FPsetca1
                      FPsetcb-5
                      FPsetcc6
                      FPqsolvexonextwocacbcc
                      FPevalxoneclip(round(xone:4))
                      FPevalxtwoclip(round(xtwo:4))
                      Quadratic equation : $ca x^2 addtermcb x addtermcc=0$\[1cm]
                      Result: $x = xone quad textand quad x = xtwo$
                      endframe
                      enddocument


                      enter image description here






                      share|improve this answer

























                      • could you explain newcommand and def...

                        – sandu
                        Mar 20 at 11:27











                      • @sandu I have edited the answer to provide context.

                        – Steven B. Segletes
                        Mar 20 at 11:32













                      6












                      6








                      6







                      Will also work with addterm -5x in addition to the intended addtermcb x.



                      The addterm macro takes a single argument, expands it once, and passes it to addtermaux. The addtermaux definition will grab the first token of the argument and examine to see if it is a minus - character. If so, it typesets a - and the rest of the argument. If not, it sees whether the first token was a + character. If so, it typesets a + and the rest of the argument. If neither of the above cases apply, it typesets a + and the complete argument.



                      In this way, the right output is provided whether cc is set to 6 or set to +6.



                      documentclassbeamer
                      usepackagefp
                      newcommandaddterm[1]expandafteraddtermaux#1relax
                      defaddtermaux#1#2relaxifx-#1-#2elseifx+#1+#2else+#1#2fifi
                      begindocument
                      beginframeQuadratic equation
                      FPsetca1
                      FPsetcb-5
                      FPsetcc6
                      FPqsolvexonextwocacbcc
                      FPevalxoneclip(round(xone:4))
                      FPevalxtwoclip(round(xtwo:4))
                      Quadratic equation : $ca x^2 addtermcb x addtermcc=0$\[1cm]
                      Result: $x = xone quad textand quad x = xtwo$
                      endframe
                      enddocument


                      enter image description here






                      share|improve this answer















                      Will also work with addterm -5x in addition to the intended addtermcb x.



                      The addterm macro takes a single argument, expands it once, and passes it to addtermaux. The addtermaux definition will grab the first token of the argument and examine to see if it is a minus - character. If so, it typesets a - and the rest of the argument. If not, it sees whether the first token was a + character. If so, it typesets a + and the rest of the argument. If neither of the above cases apply, it typesets a + and the complete argument.



                      In this way, the right output is provided whether cc is set to 6 or set to +6.



                      documentclassbeamer
                      usepackagefp
                      newcommandaddterm[1]expandafteraddtermaux#1relax
                      defaddtermaux#1#2relaxifx-#1-#2elseifx+#1+#2else+#1#2fifi
                      begindocument
                      beginframeQuadratic equation
                      FPsetca1
                      FPsetcb-5
                      FPsetcc6
                      FPqsolvexonextwocacbcc
                      FPevalxoneclip(round(xone:4))
                      FPevalxtwoclip(round(xtwo:4))
                      Quadratic equation : $ca x^2 addtermcb x addtermcc=0$\[1cm]
                      Result: $x = xone quad textand quad x = xtwo$
                      endframe
                      enddocument


                      enter image description here







                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited Mar 20 at 11:49

























                      answered Mar 20 at 11:20









                      Steven B. SegletesSteven B. Segletes

                      160k9204413




                      160k9204413












                      • could you explain newcommand and def...

                        – sandu
                        Mar 20 at 11:27











                      • @sandu I have edited the answer to provide context.

                        – Steven B. Segletes
                        Mar 20 at 11:32

















                      • could you explain newcommand and def...

                        – sandu
                        Mar 20 at 11:27











                      • @sandu I have edited the answer to provide context.

                        – Steven B. Segletes
                        Mar 20 at 11:32
















                      could you explain newcommand and def...

                      – sandu
                      Mar 20 at 11:27





                      could you explain newcommand and def...

                      – sandu
                      Mar 20 at 11:27













                      @sandu I have edited the answer to provide context.

                      – Steven B. Segletes
                      Mar 20 at 11:32





                      @sandu I have edited the answer to provide context.

                      – Steven B. Segletes
                      Mar 20 at 11:32











                      4















                      Edit: See below an improved version.




                      Note the [fragile] in beginframe. Necessary with FPifpos.



                      documentclassbeamer
                      usepackagefp
                      begindocument
                      beginframe[fragile]Quadratic equation
                      FPsetca1
                      FPsetcb-5
                      FPsetcc6
                      FPqsolvexonextwocacbcc
                      FPevalxoneclip(round(xone:4))
                      FPevalxtwoclip(round(xtwo:4))
                      FPevalbabsclip(round(abs(cb):4))
                      FPevalcabsclip(round(abs(cc):4))

                      Quadratic equation : $ca x^2$ FPifposcb $+$ else $-$ fi $babs x$ FPifposcc $+$ else $-$ fi $cabs=0$ %\[1cm]

                      Result: $x = xone quad textand quad x = xtwo$


                      endframe
                      enddocument


                      enter image description here



                      Improved version



                      This version handle better special situations (when some coefficients of the equation became -1, 1 or 0).



                      As fp's FPqsolve doesn't handle equations without solutions (it emit an error), my code don't display correctly equations where the "x^2" AND the "x" term are null (it display, when the compiler don't stop at errors, something like: Quadratic equation: +6 = 0). This code is intended to be used only when the equation has real(s) solution(s).



                      documentclassbeamer
                      usepackagefp
                      begindocument
                      beginframe[fragile]Quadratic equation
                      FPsetca1
                      FPsetcb-5
                      FPsetcc6
                      FPqsolvexonextwocacbcc
                      FPevalxoneclip(round(xone:4))
                      FPevalxtwoclip(round(xtwo:4))
                      FPevalaabsclip(round(abs(ca):4))
                      FPevalbabsclip(round(abs(cb):4))
                      FPevalcabsclip(round(abs(cc):4))
                      newcommandsignaFPifnegca -elsefi
                      newcommandpositiveSignBWithAFPifzeroca else +fi % if ca is 0, no positive sign before the "x" term if cb is positive
                      newcommandsignbFPifnegcb -else positiveSignBWithAfi
                      newcommandsigncFPifnegcc -else +fi
                      newcommandcoeffaFPifeqaabs1 elseaabsfi
                      newcommandcoeffbFPifeqbabs1 elsebabsfi
                      newcommandpolyaFPifzeroca elsesignacoeffa x^2fi
                      newcommandpolybFPifzerocb elsesignbcoeffb xfi
                      newcommandpolycFPifzerocc elsesignccabsfi

                      Quadratic equation : $polya polyb polyc =0$ \[1cm]

                      Result: $x = xone quad textand quad x = xtwo$

                      endframe
                      enddocument


                      enter image description here






                      share|improve this answer





























                        4















                        Edit: See below an improved version.




                        Note the [fragile] in beginframe. Necessary with FPifpos.



                        documentclassbeamer
                        usepackagefp
                        begindocument
                        beginframe[fragile]Quadratic equation
                        FPsetca1
                        FPsetcb-5
                        FPsetcc6
                        FPqsolvexonextwocacbcc
                        FPevalxoneclip(round(xone:4))
                        FPevalxtwoclip(round(xtwo:4))
                        FPevalbabsclip(round(abs(cb):4))
                        FPevalcabsclip(round(abs(cc):4))

                        Quadratic equation : $ca x^2$ FPifposcb $+$ else $-$ fi $babs x$ FPifposcc $+$ else $-$ fi $cabs=0$ %\[1cm]

                        Result: $x = xone quad textand quad x = xtwo$


                        endframe
                        enddocument


                        enter image description here



                        Improved version



                        This version handle better special situations (when some coefficients of the equation became -1, 1 or 0).



                        As fp's FPqsolve doesn't handle equations without solutions (it emit an error), my code don't display correctly equations where the "x^2" AND the "x" term are null (it display, when the compiler don't stop at errors, something like: Quadratic equation: +6 = 0). This code is intended to be used only when the equation has real(s) solution(s).



                        documentclassbeamer
                        usepackagefp
                        begindocument
                        beginframe[fragile]Quadratic equation
                        FPsetca1
                        FPsetcb-5
                        FPsetcc6
                        FPqsolvexonextwocacbcc
                        FPevalxoneclip(round(xone:4))
                        FPevalxtwoclip(round(xtwo:4))
                        FPevalaabsclip(round(abs(ca):4))
                        FPevalbabsclip(round(abs(cb):4))
                        FPevalcabsclip(round(abs(cc):4))
                        newcommandsignaFPifnegca -elsefi
                        newcommandpositiveSignBWithAFPifzeroca else +fi % if ca is 0, no positive sign before the "x" term if cb is positive
                        newcommandsignbFPifnegcb -else positiveSignBWithAfi
                        newcommandsigncFPifnegcc -else +fi
                        newcommandcoeffaFPifeqaabs1 elseaabsfi
                        newcommandcoeffbFPifeqbabs1 elsebabsfi
                        newcommandpolyaFPifzeroca elsesignacoeffa x^2fi
                        newcommandpolybFPifzerocb elsesignbcoeffb xfi
                        newcommandpolycFPifzerocc elsesignccabsfi

                        Quadratic equation : $polya polyb polyc =0$ \[1cm]

                        Result: $x = xone quad textand quad x = xtwo$

                        endframe
                        enddocument


                        enter image description here






                        share|improve this answer



























                          4












                          4








                          4








                          Edit: See below an improved version.




                          Note the [fragile] in beginframe. Necessary with FPifpos.



                          documentclassbeamer
                          usepackagefp
                          begindocument
                          beginframe[fragile]Quadratic equation
                          FPsetca1
                          FPsetcb-5
                          FPsetcc6
                          FPqsolvexonextwocacbcc
                          FPevalxoneclip(round(xone:4))
                          FPevalxtwoclip(round(xtwo:4))
                          FPevalbabsclip(round(abs(cb):4))
                          FPevalcabsclip(round(abs(cc):4))

                          Quadratic equation : $ca x^2$ FPifposcb $+$ else $-$ fi $babs x$ FPifposcc $+$ else $-$ fi $cabs=0$ %\[1cm]

                          Result: $x = xone quad textand quad x = xtwo$


                          endframe
                          enddocument


                          enter image description here



                          Improved version



                          This version handle better special situations (when some coefficients of the equation became -1, 1 or 0).



                          As fp's FPqsolve doesn't handle equations without solutions (it emit an error), my code don't display correctly equations where the "x^2" AND the "x" term are null (it display, when the compiler don't stop at errors, something like: Quadratic equation: +6 = 0). This code is intended to be used only when the equation has real(s) solution(s).



                          documentclassbeamer
                          usepackagefp
                          begindocument
                          beginframe[fragile]Quadratic equation
                          FPsetca1
                          FPsetcb-5
                          FPsetcc6
                          FPqsolvexonextwocacbcc
                          FPevalxoneclip(round(xone:4))
                          FPevalxtwoclip(round(xtwo:4))
                          FPevalaabsclip(round(abs(ca):4))
                          FPevalbabsclip(round(abs(cb):4))
                          FPevalcabsclip(round(abs(cc):4))
                          newcommandsignaFPifnegca -elsefi
                          newcommandpositiveSignBWithAFPifzeroca else +fi % if ca is 0, no positive sign before the "x" term if cb is positive
                          newcommandsignbFPifnegcb -else positiveSignBWithAfi
                          newcommandsigncFPifnegcc -else +fi
                          newcommandcoeffaFPifeqaabs1 elseaabsfi
                          newcommandcoeffbFPifeqbabs1 elsebabsfi
                          newcommandpolyaFPifzeroca elsesignacoeffa x^2fi
                          newcommandpolybFPifzerocb elsesignbcoeffb xfi
                          newcommandpolycFPifzerocc elsesignccabsfi

                          Quadratic equation : $polya polyb polyc =0$ \[1cm]

                          Result: $x = xone quad textand quad x = xtwo$

                          endframe
                          enddocument


                          enter image description here






                          share|improve this answer
















                          Edit: See below an improved version.




                          Note the [fragile] in beginframe. Necessary with FPifpos.



                          documentclassbeamer
                          usepackagefp
                          begindocument
                          beginframe[fragile]Quadratic equation
                          FPsetca1
                          FPsetcb-5
                          FPsetcc6
                          FPqsolvexonextwocacbcc
                          FPevalxoneclip(round(xone:4))
                          FPevalxtwoclip(round(xtwo:4))
                          FPevalbabsclip(round(abs(cb):4))
                          FPevalcabsclip(round(abs(cc):4))

                          Quadratic equation : $ca x^2$ FPifposcb $+$ else $-$ fi $babs x$ FPifposcc $+$ else $-$ fi $cabs=0$ %\[1cm]

                          Result: $x = xone quad textand quad x = xtwo$


                          endframe
                          enddocument


                          enter image description here



                          Improved version



                          This version handle better special situations (when some coefficients of the equation became -1, 1 or 0).



                          As fp's FPqsolve doesn't handle equations without solutions (it emit an error), my code don't display correctly equations where the "x^2" AND the "x" term are null (it display, when the compiler don't stop at errors, something like: Quadratic equation: +6 = 0). This code is intended to be used only when the equation has real(s) solution(s).



                          documentclassbeamer
                          usepackagefp
                          begindocument
                          beginframe[fragile]Quadratic equation
                          FPsetca1
                          FPsetcb-5
                          FPsetcc6
                          FPqsolvexonextwocacbcc
                          FPevalxoneclip(round(xone:4))
                          FPevalxtwoclip(round(xtwo:4))
                          FPevalaabsclip(round(abs(ca):4))
                          FPevalbabsclip(round(abs(cb):4))
                          FPevalcabsclip(round(abs(cc):4))
                          newcommandsignaFPifnegca -elsefi
                          newcommandpositiveSignBWithAFPifzeroca else +fi % if ca is 0, no positive sign before the "x" term if cb is positive
                          newcommandsignbFPifnegcb -else positiveSignBWithAfi
                          newcommandsigncFPifnegcc -else +fi
                          newcommandcoeffaFPifeqaabs1 elseaabsfi
                          newcommandcoeffbFPifeqbabs1 elsebabsfi
                          newcommandpolyaFPifzeroca elsesignacoeffa x^2fi
                          newcommandpolybFPifzerocb elsesignbcoeffb xfi
                          newcommandpolycFPifzerocc elsesignccabsfi

                          Quadratic equation : $polya polyb polyc =0$ \[1cm]

                          Result: $x = xone quad textand quad x = xtwo$

                          endframe
                          enddocument


                          enter image description here







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited Mar 21 at 3:26

























                          answered Mar 20 at 11:05









                          quark67quark67

                          59526




                          59526



























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