Ytdyklly

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

documentclass{beamer}

usepackage{fp}

begin{document}

begin{frame}{Quadratic equation}

FPsetca{1}

FPsetcb{-5}

FPsetcc{6}

FPqsolvexonextwocacbcc

FPevalxone{clip(round(xone:4))}

FPevalxtwo{clip(round(xtwo:4))}

Quadratic equation : $ca x^2+cb x + cc=0$\[1cm]

Result: $x = xone quad text{and} quad  x = xtwo$

end{frame}

end{document}

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

documentclass{beamer}

usepackage{fp}



newcommand{quadratic}[4][x]{%

  FPsetca{#2}%

  FPsetcb{#3}%

  FPsetcc{#4}%

  FPqsolvexonextwocacbcc

  FPevalxone{clip(round(xone:4))}%

  FPevalxtwo{clip(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 $#1=xtwo$%

}



begin{document}

begin{frame}{Quadratic equation}



quadratic{1}{-5}{6}



bigskip



quadratic[t]{2}{3}{1}



bigskip



quadratic{2}{0}{-8}



end{frame}

end{document}

With expl3:

documentclass{beamer}

usepackage{xparse}



ExplSyntaxOn



NewDocumentCommand{quadratic}{O{x}mmm}

 {

  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



begin{document}

begin{frame}{Quadratic equation}



quadratic{1}{-5}{6}



bigskip



quadratic[t]{2}{3}{1}



bigskip



quadratic{2}{0}{-8}



end{frame}

end{document}

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.

documentclass{beamer}

usepackage{fp}

newcommandaddterm[1]{expandafteraddtermaux#1relax}

defaddtermaux#1#2relax{ifx-#1-#2elseifx+#1+#2else+#1#2fifi}

begin{document}

begin{frame}{Quadratic equation}

FPsetca{1}

FPsetcb{-5}

FPsetcc{6}

FPqsolvexonextwocacbcc

FPevalxone{clip(round(xone:4))}

FPevalxtwo{clip(round(xtwo:4))}

Quadratic equation : $ca x^2 addtermcb x addtermcc=0$\[1cm]

Result: $x = xone quad text{and} quad  x = xtwo$

end{frame}

end{document}

Note the [fragile] in begin{frame}. Necessary with FPifpos.

documentclass{beamer}

usepackage{fp}

begin{document}

begin{frame}[fragile]{Quadratic equation}

FPsetca{1}

FPsetcb{-5}

FPsetcc{6}

FPqsolvexonextwocacbcc

FPevalxone{clip(round(xone:4))}

FPevalxtwo{clip(round(xtwo:4))}

FPevalbabs{clip(round(abs(cb):4))}

FPevalcabs{clip(round(abs(cc):4))}



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



Result: $x = xone quad text{and} quad  x = xtwo$





end{frame}

end{document}