Selecting cubic roots in functional form
- To: mathgroup at smc.vnet.net
- Subject: [mg50282] Selecting cubic roots in functional form
- From: suomesta at yahoo.com (Carol Ting)
- Date: Tue, 24 Aug 2004 06:22:27 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Hello, list, I have an equation 2000c*(5c+7q)-4(c+8q)*(12q+5c)^2==0 and I need to find q as a function of c, where both c and q are positive quantities. I use InequalitySolve and find that when c<3.11665 the 3rd root gives the positive function I need, and when c>3.11665 I should pick the first root instead. So I need to pick different roots in different regions of c and combine them into a piecewise continuous function. Trouble is, I have a whole series of equations generated by another function and it is impossible for me to pick the roots one by one. Is there a way to make Mathematica do this automatically? Thanks in advance! Carol
- Follow-Ups:
- Re: Selecting cubic roots in functional form
- From: DrBob <drbob@bigfoot.com>
- Re: Selecting cubic roots in functional form