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

Home > Just Math
xyz radicals (Posted on 2008-10-13) Difficulty: 3 of 5
Find all possible real triplet(s) (x,y,z) that satisfy the following system of equations:

     3 = √x + √y + √z,
     3 = x√x + y√y + z√z,
     3 = x2√x + y2√y + z2√z.

  Submitted by Bractals    
Rating: 4.0000 (1 votes)
Solution: (Hide)


Adding the first and third equations and subtracting twice the second equation gives

   3 + 3 - 2(3) = (1 + x2 - 2x)√x + (1 + y2 - 2y)√y + (1 + z2 - 2z)√z

                             or

   (x - 1)2√x + (y - 1)2√y + (z - 1)2√z = 0
Each of the three terms is nonnegative. Therefore,

   (x - 1)2√x = (y - 1)2√y = (z - 1)2√z = 0
Therefore, x, y, and z must be 0 or 1.

(1,1,1) is the only one, of the eight possible, that satisfies all three equations.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
AnswerK Sengupta2009-01-02 16:05:37
SolutionSolutionPraneeth2008-10-13 19:42:44
SolutionSolutionJonathan Lindgren2008-10-13 15:57:59
Hints/Tipsa humble startAdy TZIDON2008-10-13 12:14:16
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 (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information