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

Home > Just Math
Geometric Rubric (Posted on 2015-04-04) Difficulty: 3 of 5
The four roots of 16x4mx3+ (2m + 17)x2- mx+16 = 0 are distinct and in geometric sequence.

Determine m and solve the equation.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution re: Something cool I never knew about polynomials (hint) Comment 2 of 2 |
(In reply to Something cool I never knew about polynomials (hint) by Jer)

A good tip, permitting an easyish solution:

It was shown before somewhere that if the roots of ax^4+bx^3+cx^2+dx+e be p,q,r,s, then p+q+r+s = -b/a (in this case, m/16)
          
Applying your rule: 1/u+1/v+v+u = m/16          
          
Applying the given rule: 1/v=1/u*t, v=1/u*t^2,u=1/u*t^3, so 1/u+t/u+t^2/u+t^3/u = m/16          

If t^3/u=u, then t=u^(2/3)          
          
1/u+u^(2/3)/u+(u^(2/3))^2/u+u = m/16, equivalently u+u^(1/3)+1/u^(1/3)+1/u = m/16        

Bearing in mind those cube roots, and that integer under m, u=8 seems a pretty obvious choice.          

Then m=16*(8+2+1/2+1/8)=170 and the roots are : 1/8,1/2,2,8, with t=4
          
Checking:          
16(x-1/8)(x-1/2)(x-2)(x-8) = (8x-1)(2x-1)(x-2)(x-8) = 16x^4-(170)x^3+(2*170+17)x^2-(170)x+16, the given equation.

 

Edited on April 5, 2015, 1:42 am
  Posted by broll on 2015-04-05 01:36:32

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 (4)
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