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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

See The Solution Submitted by K Sengupta    
Rating: 3.5000 (2 votes)

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Solution computer aided solution | Comment 3 of 7 |

The curve of q^2 = p 1/p, with q as the dependent variable consists of a parabola-like figure with it's axis being the p axis (independent variable axis) and vertex at (1,0) in union with a curve going through (-1,0) asymptotically approaching the q axis towards +infinity and also asymptotically approaching the q axis towards -infinity.


The curve q = 4/(1/p + p) is valid for all real p (if we define (0,0) as part of the curve, but this doesn't play into the solution). It has negative q for negative p and positive q for positive p and is in fact what's called an odd function; q approaches zero as p becomes larger or as p becomes smaller; q is a minimum at (-1,-2) and maximum at (1,2).  It intersects the other graph in each of the two parts described above. One of the intersections has 2<p<3, 1<q<2, and the other has 1<p<0, -2<q<-1. From a graph we can see there are no other solutions, as the first equation has no values of p < -1 and values of q get farther from zero for higher values of p than 1, while values of q approach zero for larger values of p in the other equation.


The following program finds the positive solution:



p = 2.5: q = 1.4



 prevP = p: prevQ = q

 p = ((4 + SQR(16 - 4 * q * q)) / (2 * q) + p) / 2

 q = (SQR(p - 1 / p) + q) / 2

 PRINT p; q, p * p; q * q

LOOP UNTIL prevP = p AND prevQ = q


Which displays as its final iteration:


2.414213562373095  1.414213562373095      5.82842712474619  2


for p, q, p^2 and q^2.  So this solution is q=sqrt(2), p=sqrt(2) + 1, which checks out by plugging in to the equations.


By inspection, guided by the above solution and the graph, the other solution is q=-sqrt(2), p=1-sqrt(2).

Edited on July 10, 2007, 11:16 am
  Posted by Charlie on 2007-07-10 11:16:04

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