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

Home > Just Math
3 Distinct Roots Not (Posted on 2008-03-04) Difficulty: 3 of 5
Prove that the equation

   x3 + 2px2 + 2p2x + p = 0

cannot have three distinct real roots, for any real number p.

See The Solution Submitted by Bractals    
Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution | Comment 1 of 4

In order to have three real roots, the function must have a local minimum and a local maximum. For that to happen, the derivative must be zero at those points.

The derivative of x^3 + 2px^2 + 2p^2 x + p is

3x^2 + 4px + 2p^2

To solve this for zero, the discriminant is 16p^2 - 24p^2, which is negative, indicating no real roots. Therefore the derivative is never zero, there are no local minumum nor maximum, and the cubic has only one real root.

  Posted by Charlie on 2008-03-04 11:13:44
Please log in:
Remember me:
Sign up! | Forgot password

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

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information