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MathGroup Archive 2002

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RE: Coefficient problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg36435] RE: [mg36421] Coefficient problem
  • From: "DrBob" <drbob at bigfoot.com>
  • Date: Sat, 7 Sep 2002 02:53:41 -0400 (EDT)
  • Reply-to: <drbob at bigfoot.com>
  • Sender: owner-wri-mathgroup at wolfram.com

This works:

eq = a x^2 + b x + c
discriminant[eq_, x_] := Coefficient[eq, x]^2 - 4 Coefficient[eq,
   x, 2] Coefficient[eq, x, 0]
discriminant[eq, x]

x^0 is reduced to 1 and the Coefficient of 1 doesn't make sense to
Mathematica, because it depends on what the variable is (a, b, c, or
x?).  So, the other form of the Coefficient call is needed.  I used it
for the second power too, but that wasn't necessary.  I think that form
is best, though, since it allows no ambiguity.

I renamed the function because that's what the quantity is often called,
for a quadratic.

Bobby Treat

-----Original Message-----
From: CeZaR [mailto:pascal at go.ro] 
To: mathgroup at smc.vnet.net
Subject: [mg36435] [mg36421] Coefficient problem


Hi,

Now I'm trying to calculate this formula:

Delta[eq_, x_]:=Coefficient[eq, x]^2 - 4 Coefficient[eq, x^2]
Coefficient[eq, x^0]

eq has this form a x^2 + b x + c

But there is a problem with the x^0 coefficient!
How can I overcome that?

Thanks!

CeZaR





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