As Elijah pointed out, you probably are interested only in the rational
solutions, since there an infinity of real solutions, most of which are
likely to be irrational. You can even locate them. Assume you have an
integer b, such that b(b+1) > a (where a is the number you are trying to
reach – in your case 4002).
Now consider the function f: [0, 1] → R, h → (b+h)·b·h . f(0) = 0 < a,
f(1) = b(b+1) > a. It is clearly continues, strictly inscreasing,
therefore there is a unique h in (0, 1) such that (b+h)bh = a. This
means that between each integer, starting from approximately the square
root of a, there is a solution.
For instance, starting with b = 151 (taken at random), Newton's method
will quickly converge to 151.1753, which is indeed an approximation of a
solution.
Adrien Mathieu
On 19/10/2021 11:08, Elijah Stone wrote:
On Tue, 19 Oct 2021, Elijah Stone wrote:
0 = (64*x*x) + (_4096*x) + 4002
Another foolish mistake! It should be '- 4002', not '+ 4002'. (The
answer is still correct; but I did not correctly reproduce my workings.)
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