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find the root of the function

🔗nasnas_100 <nasnas_100@yahoo.co.uk>

8/9/2002 2:21:41 PM

hi i am new student i have a problem
find the root of f(x)=x^3+30x-30
thanks

🔗wallyesterpaulrus <perlich@aya.yale.edu>

8/9/2002 3:06:40 PM

--- In tuning-math@y..., "nasnas_100" <nasnas_100@y...> wrote:

> hi i am new student i have a problem
> find the root of f(x)=x^3+30x-30
> thanks

what does this have to do with tuning?

🔗M. Edward Borasky <znmeb@aracnet.com>

8/10/2002 1:59:06 PM

Well . according to Derive, there are two complex roots and one real root:

x = -0.4848069410 - 5.541219478.i ; x = -0.4848069410 + 5.541219478.i ; x
= 0.9696138820

Now that I've given you the answer, your assignment is to look up the
formula (and there is one) for solving a cubic (which this is) equation and
post it to the list. Then tell us what this has to do with music / tuning.

-----Original Message-----
From: nasnas_100 [mailto:nasnas_100@yahoo.co.uk]
Sent: Friday, August 09, 2002 2:22 PM
To: tuning-math@yahoogroups.com
Subject: [tuning-math] find the root of the function

hi i am new student i have a problem
find the root of f(x)=x^3+30x-30
thanks

🔗Gene W Smith <genewardsmith@juno.com>

8/10/2002 5:58:22 PM

On Sat, 10 Aug 2002 13:59:06 -0700 "M. Edward Borasky"
<znmeb@aracnet.com> writes:
> Well . according to Derive, there are two complex roots and one real
> root:
>
> x = -0.4848069410 - 5.541219478.i ; x = -0.4848069410 +
> 5.541219478.i ; x
> = 0.9696138820
>
> Now that I've given you the answer, your assignment is to look up the
> formula (and there is one)

There's more than one, and solving it in radicals (as opposed to
Chebyshev radicals, for instance) isn't the neatest.