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 A Reciprocal And Square Problem (Posted on 2007-07-10)
Find all real pairs (p, q) satisfying the following system of equations:
p - 1/p - q2 = 0

q/p + pq = 4

 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

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