Re: How can I solve these simultaneous quadratic equations in Mathemetica?

*To*: mathgroup at smc.vnet.net*Subject*: [mg65775] Re: How can I solve these simultaneous quadratic equations in Mathemetica?*From*: "Chris Chiasson" <chris.chiasson at gmail.com>*Date*: Mon, 17 Apr 2006 02:28:16 -0400 (EDT)*References*: <e1stfp$bf5$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Thoihen (cool name), There are some ways to get some solutions to your system. I am not sophistocated at math (or spelling), but here are two solutions, one at c=14/10 and the other at c=1/2: eqns={1+3c/2+x(1+c+y+z-x)+c*y+(y+z+x)/2-w(2y+2x-3w+3+2c)\[Equal]0, c/2+w(y+z-w+3/2+c)-x(2w+2z-3x+2)\[Equal]0, 1+2c+w(1+c)+z(w+x-z+3/2+c)-y(2w+2z-3y+4+2c)\[Equal]0, y(x+w-y+1+c)-c(1+w)+(y+w-x)/2-z(2y+2x-3z)\[Equal]0} FindRoot[eqns/.c\[Rule]1/2,{{w,2},{x,3},{y,4},{z,5}}] FindRoot[eqns/.c\[Rule]14/10,{{w,2},{x,3},{y,4},{z,5}}] One way to search for solutions (assuming the starting guesses are sufficient) is to change the value of c and look at the residuals. In this problem, it seems as if FindRoot will warn when its results won't be correct. At first I just made a table and changed the step size for c until I got some solutions. Eventually, I changed it to a graphing routine. If you could provide more information about what these parameters represent and how you came to these equations, I am sure someone could suggest a better method to go about finding solutions. (for the code below: where the residuals go to zero, you may find a good solution) gendat=Table[{c, Evaluate[eqns]/.Equal\[Rule]Subtract/.FindRoot[ Evaluate[eqns],{{w,2},{x,3},{y,4},{z,5}}]},{c,0,2,0.001}]; gendatplotdat= Transpose[Map[Function[{datrow},{datrow[[1]],#}&/@datrow[[2]]],gendat]]; Show[Block[{$DisplayFunction=Identity}, MapIndexed[ ListPlot[#1,PlotJoined\[Rule]True, PlotStyle\[Rule]Hue[#2[[1]]/Length[gendatplotdat]]]&, gendatplotdat]],PlotRange\[Rule]All,ImageSize\[Rule]72*8]