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A Reciprocal And Square Problem (Posted on 2007-07-10) Difficulty: 2 of 5
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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: Analytical solution - correctedAdy TZIDON2007-07-11 07:19:10
SolutionSolPraneeth Yalavarthi2007-07-10 11:44:22
SolutionAnalytical solutionFederico Kereki2007-07-10 11:38:51
re(2): SolutionPraneeth Yalavarthi2007-07-10 11:30:14
Solutioncomputer aided solutionCharlie2007-07-10 11:16:04
re: SolutionCharlie2007-07-10 11:08:38
SolutionPraneeth Yalavarthi2007-07-10 10:15:27
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