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

Home > Just Math
Supreme Value Validation (Posted on 2013-09-08) Difficulty: 3 of 5
P1(x) = x2 + (k-29)x - k and, P2(x) = 2x2+(2k-43)x + k, where each of P1(x) and P2(x) is a factor of a cubic polynomial P(x).

Determine the maximum value of k.

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.)
Solution Another approach Comment 3 of 3 |
As Daniel noted, P1 and P2 must have a common root.  That root can be found by calculating 2*P1(x) - P2(x) = 2*(x^2 + (k-29)x - k) -(2x^2+(2k-43)x + k) = -15x - 3k = -15*(x+k/5).

x = -k/5 is the common root of P1 and P2.  Plug that in and equate the equations to zero: (-k/5)^2 + (k-29)*(-k/5) - k = 0 and 2*(-k/5)^2 + (2k-43)*(-k/5) + k = 0.

The two equations reduce to (-4/25)k^2 + (24/5)k = 0 and (-8/25)k^2 + (48/5)k = 0 respectively.  These are the same quadratic, just one has coefficients twice as large.  The roots are k=30 and k=0.

Then the total set of k such that P1 and P2 are factors of the cubic P(x) are {0,30}.  The maximum value of k must be 30.

  Posted by Brian Smith on 2017-01-01 18:52:39
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