Contribution for Puzzlet #071
From:  Ken Duisenberg []

From the problem statement, we know that
c = -(92^2-b^2)/4a = ([b/2]^2 -

For c to be positive, b >= 94, and even, and
we already know b<100, so there
are only
three values to check.

b=94 -> c<100, so doesn't work.
b=96, c=188, a=1 (any higher a and c < 100)
b=98, c=285, a=1 (any higher a and c < 100)

Thanks for the response.  Your approach is very neat, and, of course, generates the correct constants, which would lead to the roots of  (-94, -2) and (-95, -3) for the two sets of answers.  I was being a bit fastidious when I said "by hand" since I was thinking of eliminating any trial and error at all, but you've pointed out that it can only be two out of three trials anyway - so hats off to you. The Puzzlet will still entertain those who prefer to solve these things by devising an algorithm and writing some code.



Site design/maintenance: Dave Ellis E-mail me!
Last Updated: February 22nd, 2004.