Re: transposing an equation
- To: mathgroup at smc.vnet.net
- Subject: [mg26294] Re: transposing an equation
- From: leko at ix.netcom.com (J. Leko)
- Date: Sun, 10 Dec 2000 00:20:04 -0500 (EST)
- References: <90kq3d$r25@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
You could do this:
In[2]:= Solve[y == a x^2 + b x + c, x]
Out[2]=
\!\({{x -> \(\(-b\) + \ at \(b\^2 - 4\ a\ c + 4\ a\ y\)\)\/\(2\ a\)}, {x -> \
\(-\(\(b + \ at \(b\^2 - 4\ a\ c + 4\ a\ y\)\)\/\(2\ a\)\)\)}}\)
(sorry about the formating). Although in this case, you get something
similar to the quadratic equation for answers, so there are really no A,
B, and C's.
In article <90kq3d$r25 at smc.vnet.net>, "Christopher Deacon"
<cdeacon at physics.mun.ca> wrote:
> Suppose y=a x^2+b x +c.
>
> How can I get Mathematica to solve for x (i.e., x=Ay^2+By+C) and give me the
> values for the constants A,B,C?
>
> Chris
>
> --
> +-----------------------------+----------------------------+
> | Christopher Deacon | (709) 737-7631
> | Dept of Physics and Physical| cdeacon at physics.mun.ca
> | Oceanography
> | Memorial University of Nfld
> +----------------------------+-----------------------------+
> | http://www.physics.mun.ca/~cdeacon
> +----------------------------------------------------------+
J. Leko
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