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Take Third Degree, Solve For Real (Posted on 2007-06-27) Difficulty: 3 of 5
Determine all possible real pairs (p, q) satisfying:

2q3 – p3 = pq2+ 11

See The Solution Submitted by K Sengupta    
Rating: 3.0000 (2 votes)

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Some Thoughts some thoughts | Comment 1 of 4

I had the idea to introduce a parameter into the problem to see if I could simplify it.  We are looking for real pairs (p,q) that satisfy the given equation.  So, letting q=kp (where k is some unknown real variable, the parameter):

2q^3 - p^3 = pq^2 + 11

2(kp)^3 - p^3 = p(kp)^2 + 11

2(k^3)(p^3) - p^3 = (k^2)(p^3) + 11

(p^3)[2k^3 - k^2 - 1] = 11

p = [11/(2k^3 - k^2 - 1)]^(1/3)

This means that all real solutions to the given equation are expressible in the form

(p,q) = ([11/(2k^3 - k^2 - 1)]^(1/3) , k*[11/(2k^3 - k^2 - 1)]^(1/3))

Here, the parameter k can range over all possible real numbers except 1 (if we let k = 1, we get division by zero in the expressions given above).  [Note: Also, we have to include the point

(0, (11/2)^(1/3))

which is the limiting value of the parametric expression as we let k approach positive or negative infinity.]

I was wondering what this solution curve looks like!  I was trying to work out what it would look like with pencil and paper, but I might have to resort to the graphing calculator here...

Any other ideas on this problem?  It's a good one, thanks K Sengupta!

-John

ps  Alas, I have been foiled by the math symbols once again!  I'm editing this comment so that it will hopefully be more readable.

Edited on June 27, 2007, 8:47 pm
  Posted by John Reid on 2007-06-27 17:20:44

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