HP Forum Archive 14

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33s solving symmetrical quadratics Message #1 Posted by bill platt on 28 July 2004, 9:45 a.m. Hi, Curious business. Try solving the simple expression n^2-6 on the 33s in the equation list. It will return the positive root, but not the negative. Try controlling the starting estimates by storing -10 to n, and -1 in the x-register. Still, a positive root. Try storing -1 in n, and -10 in the x-register. still, only a positive root! On the 32sii, controlling the starting estimates "kicks" the solution over to the negative root. Try a parabola that is shifted on the x-axis: v^2-v-6. When you play with the initial estimates on this one, you DO get the negative root. Strange. I wonder if this is related to the bizarrre "-0" issue which crops up in the cartesian to polar conversion issue.? Regards, Bill

Re: 33s solving symmetrical quadratics Message #2 Posted by bill platt on 28 July 2004, 2:08 p.m., in response to message #1 by bill platt But, If you solve x^2 - 6 in either: a program (RPN or ALG) a program using an equation You *can* push the solution across the zero to find the negative root. Hmmmmm..... Regards, Bill Edited: 28 July 2004, 2:09 p.m.

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