|33s solving symmetrical quadratics|
Message #1 Posted by bill platt on 28 July 2004, 9:45 a.m.
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.?