2 Feb
2007
2 Feb
'07
11:39 a.m.
About x(x² + a)(x² + b) = c Thanks Joshua and Fred for your first answers. I know these old methods. By sometimes better suitable methods exist depending on the equation to analyze. a, b, c can greatly vary, but are large numbers. Not close to 1. b is approx. of the same size than a, say: a < b < 12a. And because x(x²+a)(x²+b) is a growing function, we are sure that the solution x exists, is real, and is unique. Franck, you are perfectly right, the key is to find a good starting point. What would be a good starting point? Christian.