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

Home > Just Math
Get The Quadruplets (Posted on 2007-04-06) Difficulty: 2 of 5
Analytically determine all possible quadruplets (p, q, r, s) of real numbers satisfying the following system of equations:

p+q = 8
pq + r + s = 23
ps + qr = 28
rs = 12

See The Solution Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution All Answers Analytically Comment 3 of 3 |

From the given eequations, let s = 12/r and q = 8-p:
p(8-p) + r + 12/r = 23
12*p/r + (8-p)r = 28

Rearrange the second new equation as:
12p - pr^2 = 28r - 8r^2
p = (28r - 8r^2)/(12 - r^2)

Substituting the expression for p into the p/r equation:
(28r - 8r^2)/(12 - r^2) * (8-(28r - 8r^2)/(12 - r^2)) + r + 12/r = 23

r*(28r - 8r^2)*(8*(12 - r^2) - (28r - 8r^2)) + (r^2+12)*(12 - r^2)^2 = 23*r*(12 - r^2)^2

(28r^2 - 8r^3)*(96 - 28r) + (144 - r^4)*(12 - r^2) = 23r*(r^4 - 24r^2 + 144)

(2688r^2 - 768r^3 - 784r^3 + 224r^4) + (1728 - 144r^2 - 12r^4 + r^6) - (23r^5 - 552r^3 + 3312r) = 0

r^6 - 23r^5 + 212r^4 - 1000r^3 + 2544r^2 - 3312r + 1728 = 0

Solving this polynomial yields roots of: {2, 2, 3, 4, 6, 6}
This means that there are exactly four solutions:
(p,q,r,s) = (3,5,2,6), (4,4,3,4), (4,4,4,3), (5,3,6,2)


  Posted by Brian Smith on 2007-04-09 14:28:21
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 (3)
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