All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Product = Sum + 2 (Posted on 2010-09-14) Difficulty: 3 of 5
Determine all possible triplet(s) (p, q, r) of positive real numbers, with p ≤ q ≤ r, that satisfy the following system of equations:

p*q + q*r + p*r = 12 , and:
p*q*r = p + q + r + 2

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: solution Comment 3 of 3 |
(In reply to solution by Charlie)

Rereading the puzzle, I only now realize it asks for real numbers rather than integers only. But refining the search still appears to show that the 12 equality holds only at (2,2,2) for positive reals.

I realized this after Steve Herman found a non-integral negative solution.


  Posted by Charlie on 2010-09-15 09:51:29
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (24)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information