33s solving symmetrical quadratics

[bill platt]

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

[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.