Path: news.mathworks.com!newsfeed-00.mathworks.com!solaris.cc.vt.edu!news.vt.edu!news.glorb.com!postnews.google.com!l35g2000vba.googlegroups.com!not-for-mail
From: Greg Heath <heath@alumni.brown.edu>
Newsgroups: comp.soft-sys.matlab
Subject: Re: root finding of Linear equation
Date: Sat, 21 Nov 2009 10:27:01 -0800 (PST)
Organization: http://groups.google.com
Lines: 39
Message-ID: <07a2085d-0b34-4269-9b52-b9901df3ee4a@l35g2000vba.googlegroups.com>
References: <hcbncl$aov$1@fred.mathworks.com> <hcbqad$3s7$1@fred.mathworks.com> 
	<727cae84-3988-4ae8-aaa7-cc73d80056e3@m33g2000vbi.googlegroups.com> 
	<he8cne$pvb$1@fred.mathworks.com>
NNTP-Posting-Host: 69.141.163.135
Mime-Version: 1.0
Content-Type: text/plain; charset=ISO-8859-1
Content-Transfer-Encoding: quoted-printable
X-Trace: posting.google.com 1258828021 19476 127.0.0.1 (21 Nov 2009 18:27:01 GMT)
X-Complaints-To: groups-abuse@google.com
NNTP-Posting-Date: Sat, 21 Nov 2009 18:27:01 +0000 (UTC)
Complaints-To: groups-abuse@google.com
Injection-Info: l35g2000vba.googlegroups.com; posting-host=69.141.163.135; 
	posting-account=mUealwkAAACvQrLWvunjg50tRAnsNtJR
User-Agent: G2/1.0
X-HTTP-UserAgent: Mozilla/4.0 (compatible; MSIE 7.0; Windows NT 5.1; .NET CLR 
	2.0.50727; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe)
Xref: news.mathworks.com comp.soft-sys.matlab:586904


On Nov 21, 4:43 am, "John D'Errico" <woodch...@rochester.rr.com>
wrote:
> Greg Heath <he...@alumni.brown.edu> wrote in message <727cae84-3988-4ae8-aaa7-cc73d8005...@m33g2000vbi.googlegroups.com>...
> > On Oct 29, 5:21?am, "John D'Errico" <woodch...@rochester.rr.com>
> > wrote:
> > > "Eldar " <el_os...@mynet.com> wrote in message <hcbncl$ao...@fred.mathworks.com>...
> > > > I need to find out the roots of an any linear equation like;
> > > > y = ax? + bx? + cx + d
> > > > or
> > > > y = ax? + bx + c
>
> > > > Any suggestions on how to approach such problem?
> > > > Consider that I'm just new user of Matlab.
>
> > > In "ax? + bx? + cx + d", what is the ?
> > > supposed to do?
>
> > > If these were meant to indicate a power
> > > operation, then how is it that you call this
> > > a LINEAR equation? It is nonlinear in the
> > > variable of interest.
>
> > It's linear in the unknown coefficients which,
> > given data, can be solved using backslash.
>
> > Greg
>
> NO. You can solve for the coefficients using
> backslash.
>
> But you cannot solve for the roots of an equation
> using backslash!

 John, please splash your face with water.
I said given data, not given coefficients.

Greg