Take Third Degree, Solve For Real (Posted on 2007-06-27)

	Submitted by K Sengupta
Rating: 3.0000 (2 votes)
Solution:	(Hide)
At the outset, we observe that : 2q^3 – p^3 – pq^2 = (q-p)(2q^2+ pq+ q^2) = 11. Now, 2q^2 + pq+ p^2 > q-p, and accordingly, we must have: Also, 2q^2 + pq+ p^2 = (p+ q/2)^2 + (7/4)*q^2 > = 0 If, 2q^2 + pq+ p^2 = 0, then we must have: (p + q/2, q/2) = (0, 0) giving: (p, q) = 0, and accordingly, 0 divides 11. This is a contradiction. Therefore, 2q^2 + pq+ p^2 > 0, so that : q-p = 1 and 2q^2 + pq+ p^2 = 11 Accordingly, q= p+1, so that 2q^2 + pq+ p^2 = 11 gives: (p-1)(4p-9) = 0 upon simplification. p=1, gives q= 2; while p= 9/4 gives q= 13/4 Thus, (p, q) = (1,2); (9/4, 13/4) are the only possible solutions to the given problem.

	Subject	Author	Date
	No Subject	K Sengupta	2023-08-02 22:14:40
	Reader	Nesronke Hersonke	2020-11-12 00:36:56
	Factoring Idea	Brian Smith	2007-06-27 23:39:53
	some thoughts	John Reid	2007-06-27 17:20:44