A Reciprocal And Square Problem (Posted on 2007-07-10)

	Submitted by K Sengupta
Rating: 3.5000 (2 votes)
Solution:	(Hide)
The given equations yield; p – 1/p = q^2 p+ 1/p = 4/q Thus, 16/(q^2) – q^4 = 4, giving: u^3 + 4u – 16 = 0, where u = q^2 Or, (u-2)(u^2 +2u +8) = 0 Or, u =2, ignoring the complex roots of u which are inadmissible. or, q = +/- √2) If q = √2, then p = (1/2)*(q^2 + 4/q) = 1+ √2 Similarly, q = - √2 gives p = 1 – √2 Thus (p, q) = (√2, 1 + √2), (-√2, 1- √2) are the only possible solutions.

	Subject	Author	Date
	re: Analytical solution - corrected	Ady TZIDON	2007-07-11 07:19:10
	Sol	Praneeth Yalavarthi	2007-07-10 11:44:22
	Analytical solution	Federico Kereki	2007-07-10 11:38:51
	re(2): Solution	Praneeth Yalavarthi	2007-07-10 11:30:14
	computer aided solution	Charlie	2007-07-10 11:16:04
	re: Solution	Charlie	2007-07-10 11:08:38
	Solution	Praneeth Yalavarthi	2007-07-10 10:15:27