ed84c Posted October 31, 2007 Share Posted October 31, 2007 Ok so I need to solve; [math] z^{5} + x^{4} + z^{3}+z^{2}+z + 1 = 0 [/math] My first thought is to use the summation equation where a = 1, i.e.; [math] \frac {1-z^5}{1-z} = 0 [/math] but then I'm not sure where to go from there; in the past I have had something^something else = 1 then something = alpha and then made z = f(alpha) however here there isnt a whole term to the same index. Any ideas? Link to comment Share on other sites More sharing options...
timo Posted October 31, 2007 Share Posted October 31, 2007 My first idea is that three times an odd power of z plus two times an even power of z (I assume x^4 was a typo) shall equal -1 ... Link to comment Share on other sites More sharing options...
D H Posted October 31, 2007 Share Posted October 31, 2007 First thing: That's [math]\frac{1-z^6}{1-z}=0[/math], not [math]\frac{1-z^5}{1-z}=0[/math]. Multiple both sides by the denominator. [math]1-x^6=0[/math] has six solutions, one of which results in 0/0. The remaining five solutions are the five solutions to the original problem. Link to comment Share on other sites More sharing options...
ed84c Posted October 31, 2007 Author Share Posted October 31, 2007 Yep, that checks out. Cheers for that. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now