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

Home > Just Math
One way out (Posted on 2019-09-09) Difficulty: 2 of 5
f(x) is a 6th degree polynomial satisfying f(1-x)=f(x+1) for all real values of x. If f(x) has four distinct real roots and two real and equal roots, then find the sum of all roots of f(x)=0.

No Solution Yet Submitted by Danish Ahmed Khan    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution Comment 1 of 1
The function is symmetrical about the line x=1. For example f(2) = f(0), f(5) = f(-3), f(1) = f(1). What confuses things is the use of x in the condition: when that x is zero the argument to f is 1, which is traditionally the x value; it's a different x.

Since it's symmetric about x=1, first of all the two equal roots are at x=1. As far as the total goes, the easiest way to that is via the average. The average of the roots must be 1. As there are 5 roots, the total is 5. If you were to count the double root twice, the total would be 6.

  Posted by Charlie on 2019-09-09 09:29:05
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 (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2020 by Animus Pactum Consulting. All rights reserved. Privacy Information