Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2004
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Expanding a Square Root

  • To: mathgroup at smc.vnet.net
  • Subject: [mg50820] Re: [mg50813] Expanding a Square Root
  • From: Andrzej Kozlowski <andrzej at akikoz.net>
  • Date: Wed, 22 Sep 2004 02:27:37 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

On 22 Sep 2004, at 13:11, Jim Dars wrote:

> Hi All,
>
> I should point out that I use Mathematica 4.0 infrequently, thus I may 
> be
> missing the obvious.  My problem is as follows:
>
> In the following I use the square root sign from the palette for SQRT
> Expand[(44+9*SQRT(23))^2]  yields
> 3799 + 792*SQRT(23)
>
> I now wish to take the square root of  3799 + 792*SQRT(23) to return 
> to my
> original expression.  However
>
> Expand[SQRT(3799 + 792*SQRT(23))] merely yields
>
> [SQRT(3799 + 792*SQRT(23))]
>
> What is the proper command to make SQRT(3799 + 792*SQRT(23)) yield
> 44+9*SQRT(23)
>
> Best wishes, Jim
>
>
>

RootReduce[Sqrt[3799 + 792*Sqrt[23]]]


44 + 9*Sqrt[23]

You can of course also use FullSimplify instead of RootReduce but 
FullSimplify just calls RootReduce to do the job.

Andrzej Kozlowski


Chiba, Japan
http://www.akikoz.net/~andrzej/
http://www.mimuw.edu.pl/~akoz/


  • Prev by Date: Re: Telling Mathematica that a symbol is going to be a List?
  • Next by Date: Re: Expanding a Square Root
  • Previous by thread: Re: Expanding a Square Root
  • Next by thread: Telling Mathematica that a symbol is going to be a List?