How to write Quadratic equation with negative coefficientdef 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”
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How to write Quadratic equation with negative coefficient
def 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”
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
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
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
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
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
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
fp
edited Mar 21 at 3:52
sandu
asked Mar 20 at 9:54
sandusandu
3,70342856
3,70342856
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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
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
.
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
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
add a comment |
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
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
add a comment |
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3 Answers
3
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.
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
add a comment |
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
add a comment |
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
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
edited Mar 20 at 11:44
answered Mar 20 at 11:27
egregegreg
728k8819253235
728k8819253235
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
.
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
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
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
.
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
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
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
.
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
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
edited Mar 20 at 11:49
answered Mar 20 at 11:20
Steven B. SegletesSteven B. Segletes
159k9204412
159k9204412
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
add a comment |
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
add a comment |
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
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
add a comment |
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
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
add a comment |
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
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
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
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
edited Mar 21 at 3:26
answered Mar 20 at 11:05
quark67quark67
43526
43526
add a comment |
add a comment |
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