How to write Quadratic equation with negative coefficient
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}
fp
add a comment |
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}
fp
add a comment |
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}
fp
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}
fp
fp
edited 16 hours ago
sandu
asked yesterday
sandusandu
3,67942856
3,67942856
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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}

add a comment |
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}

could you explain newcommand and def...
– sandu
yesterday
@sandu I have edited the answer to provide context.
– Steven B. Segletes
yesterday
add a comment |
Edit: See below an improved version.
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}

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).
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))}
FPevalaabs{clip(round(abs(ca):4))}
FPevalbabs{clip(round(abs(cb):4))}
FPevalcabs{clip(round(abs(cc):4))}
newcommand{signa}{FPifnegca -elsefi}
newcommand{positiveSignBWithA}{FPifzeroca else +fi} % if ca is 0, no positive sign before the "x" term if cb is positive
newcommand{signb}{FPifnegcb -else positiveSignBWithAfi}
newcommand{signc}{FPifnegcc -else +fi}
newcommand{coeffa}{FPifeqaabs1 elseaabsfi}
newcommand{coeffb}{FPifeqbabs1 elsebabsfi}
newcommand{polya}{FPifzeroca elsesignacoeffa x^2fi}
newcommand{polyb}{FPifzerocb elsesignbcoeffb xfi}
newcommand{polyc}{FPifzerocc elsesignccabsfi}
Quadratic equation : $polya polyb polyc =0$ \[1cm]
Result: $x = xone quad text{and} quad x = xtwo$
end{frame}
end{document}

add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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}

add a comment |
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}

add a comment |
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}

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}

edited yesterday
answered yesterday
egregegreg
728k8819233233
728k8819233233
add a comment |
add a comment |
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}

could you explain newcommand and def...
– sandu
yesterday
@sandu I have edited the answer to provide context.
– Steven B. Segletes
yesterday
add a comment |
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}

could you explain newcommand and def...
– sandu
yesterday
@sandu I have edited the answer to provide context.
– Steven B. Segletes
yesterday
add a comment |
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}

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}

edited yesterday
answered yesterday
Steven B. SegletesSteven B. Segletes
159k9204411
159k9204411
could you explain newcommand and def...
– sandu
yesterday
@sandu I have edited the answer to provide context.
– Steven B. Segletes
yesterday
add a comment |
could you explain newcommand and def...
– sandu
yesterday
@sandu I have edited the answer to provide context.
– Steven B. Segletes
yesterday
could you explain newcommand and def...
– sandu
yesterday
could you explain newcommand and def...
– sandu
yesterday
@sandu I have edited the answer to provide context.
– Steven B. Segletes
yesterday
@sandu I have edited the answer to provide context.
– Steven B. Segletes
yesterday
add a comment |
Edit: See below an improved version.
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}

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).
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))}
FPevalaabs{clip(round(abs(ca):4))}
FPevalbabs{clip(round(abs(cb):4))}
FPevalcabs{clip(round(abs(cc):4))}
newcommand{signa}{FPifnegca -elsefi}
newcommand{positiveSignBWithA}{FPifzeroca else +fi} % if ca is 0, no positive sign before the "x" term if cb is positive
newcommand{signb}{FPifnegcb -else positiveSignBWithAfi}
newcommand{signc}{FPifnegcc -else +fi}
newcommand{coeffa}{FPifeqaabs1 elseaabsfi}
newcommand{coeffb}{FPifeqbabs1 elsebabsfi}
newcommand{polya}{FPifzeroca elsesignacoeffa x^2fi}
newcommand{polyb}{FPifzerocb elsesignbcoeffb xfi}
newcommand{polyc}{FPifzerocc elsesignccabsfi}
Quadratic equation : $polya polyb polyc =0$ \[1cm]
Result: $x = xone quad text{and} quad x = xtwo$
end{frame}
end{document}

add a comment |
Edit: See below an improved version.
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}

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).
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))}
FPevalaabs{clip(round(abs(ca):4))}
FPevalbabs{clip(round(abs(cb):4))}
FPevalcabs{clip(round(abs(cc):4))}
newcommand{signa}{FPifnegca -elsefi}
newcommand{positiveSignBWithA}{FPifzeroca else +fi} % if ca is 0, no positive sign before the "x" term if cb is positive
newcommand{signb}{FPifnegcb -else positiveSignBWithAfi}
newcommand{signc}{FPifnegcc -else +fi}
newcommand{coeffa}{FPifeqaabs1 elseaabsfi}
newcommand{coeffb}{FPifeqbabs1 elsebabsfi}
newcommand{polya}{FPifzeroca elsesignacoeffa x^2fi}
newcommand{polyb}{FPifzerocb elsesignbcoeffb xfi}
newcommand{polyc}{FPifzerocc elsesignccabsfi}
Quadratic equation : $polya polyb polyc =0$ \[1cm]
Result: $x = xone quad text{and} quad x = xtwo$
end{frame}
end{document}

add a comment |
Edit: See below an improved version.
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}

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).
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))}
FPevalaabs{clip(round(abs(ca):4))}
FPevalbabs{clip(round(abs(cb):4))}
FPevalcabs{clip(round(abs(cc):4))}
newcommand{signa}{FPifnegca -elsefi}
newcommand{positiveSignBWithA}{FPifzeroca else +fi} % if ca is 0, no positive sign before the "x" term if cb is positive
newcommand{signb}{FPifnegcb -else positiveSignBWithAfi}
newcommand{signc}{FPifnegcc -else +fi}
newcommand{coeffa}{FPifeqaabs1 elseaabsfi}
newcommand{coeffb}{FPifeqbabs1 elsebabsfi}
newcommand{polya}{FPifzeroca elsesignacoeffa x^2fi}
newcommand{polyb}{FPifzerocb elsesignbcoeffb xfi}
newcommand{polyc}{FPifzerocc elsesignccabsfi}
Quadratic equation : $polya polyb polyc =0$ \[1cm]
Result: $x = xone quad text{and} quad x = xtwo$
end{frame}
end{document}

Edit: See below an improved version.
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}

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).
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))}
FPevalaabs{clip(round(abs(ca):4))}
FPevalbabs{clip(round(abs(cb):4))}
FPevalcabs{clip(round(abs(cc):4))}
newcommand{signa}{FPifnegca -elsefi}
newcommand{positiveSignBWithA}{FPifzeroca else +fi} % if ca is 0, no positive sign before the "x" term if cb is positive
newcommand{signb}{FPifnegcb -else positiveSignBWithAfi}
newcommand{signc}{FPifnegcc -else +fi}
newcommand{coeffa}{FPifeqaabs1 elseaabsfi}
newcommand{coeffb}{FPifeqbabs1 elsebabsfi}
newcommand{polya}{FPifzeroca elsesignacoeffa x^2fi}
newcommand{polyb}{FPifzerocb elsesignbcoeffb xfi}
newcommand{polyc}{FPifzerocc elsesignccabsfi}
Quadratic equation : $polya polyb polyc =0$ \[1cm]
Result: $x = xone quad text{and} quad x = xtwo$
end{frame}
end{document}

edited 16 hours ago
answered yesterday
quark67quark67
43526
43526
add a comment |
add a comment |
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