Re: square root denominators

From: Timothy Little (tim-via-n.i.net_at_little-possums.net)
Date: 02/25/05 

Date: 25 Feb 2005 05:08:12 GMT

Oscar Lanzi III wrote:
> I had to solve by computer a quadratic equation of the form
> a^2 + bx + c = 0 with a being potentially zero. Using a
> rationalized-numerator version of the formula,
> c/(-b (+/-) sqrt(b^2-4ac)), allows one to avoid a zero or near-zero
> denominator in such a case.

If a=0, then (-b (+/-) sqrt(b^2-4ac)) is either -2b or 0. I can see
that in half the cases you avoid zero or near-zero denominator, but
what about the other half? For sufficiently small a, there will
always be two roots and one will be very large in magnitude.

> Another example is integrating sqrt((1+x)/(1-x)) dx by first
> rendering the integrand as (1+x)/sqrt(1-x^2).

That one does seem more useful.

- Tim