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
)
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.
)
A negative solution
| Comment 2 of 3 |
I know the problem asks for a positive solution, but here's a negative one: (-5.5, -1, -1)
p*q + q*r + p*r = 5.5 + 5.5 + 1 = 12
p*q*r = -.5.5 = p + q + r + 2
Posted by
Steve Herman
on 2010-09-14 22:44:56
Please log in:
Login:
Password:
Remember me:
Sign up!
|
Forgot password
Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ
|
About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
blackjack
flooble's webmaster puzzle
Copyright © 2002 - 2024 by
Animus Pactum Consulting
. All rights reserved.
Privacy Information