X^n+ax+b=0 | Free Math Help Forum

• Home
• Forums New posts Search forums
• What's new New posts Latest activity

Log in Register What's new Search

Search

Everywhere Threads This forum This thread Search titles only By: Search Advanced search…

• New posts
• Search forums

Menu Log in Register Install the app Install

• Forums
• Free Math Help
• Calculus

You are using an out of date browser. It may not display this or other websites correctly.You should upgrade or use an alternative browser. x^n+ax+b=0

• Thread starter thangdmc113
• Start date Oct 24, 2019

T

thangdmc113

New member

Joined Oct 24, 2019 Messages 1 Please help me Prove that x^n+ax+b=0 has at most 2 solution if n is odd and has at most 3 solution if n is even. Thanks D

Deleted member 4993

Guest

thangdmc113 said: Please help me Prove that x^n+ax+b=0 has at most 2 solution if n is odd and has at most 3 solution if n is even. Thanks Click to expand...

Please share your work/thoughts about this assignment. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING

Steven G

Elite Member

Joined Dec 30, 2014 Messages 14,603 I do not accept this problem as being true at all. n= 3 is an odd number but I do not believe that a cubic equation of this form can have at most 2 roots. Just graph the cubic equation y = y = x^3-5x-1 and see how many roots it has. Also a polynomial of odd degree can never have two real roots as complex roots comes in pairs. So suppose the degree of a polynomial is 7. The the number of complex roots can be 0 (and then there are 7 real roots) or the number complex roots are 2 (so there are 5 real roots) or the number of complex roots are 4 (and there are 3 real roots) or the number of complex roots are 6 (and there is 1 real root) So (in theory) it is not wrong to say that there are at most 2 real roots in a polynomial of odd degree of the form x^n +ax+b it is more exact to say that there is 1 real roots in a polynomial of odd degree of the form x^n +ax+b. The problem is that many odd degree polynomials of the form y = x^n +ax+b have more than 1 root. Is there a problem as well with the statement of even degree polynomials? Last edited: Oct 24, 2019

Dr.Peterson

Elite Member

Joined Nov 12, 2017 Messages 16,873

thangdmc113 said: Please help me Prove that x^n+ax+b=0 has at most 2 solution if n is odd and has at most 3 solution if n is even. Thanks Click to expand...

Have you omitted some conditions? For example, must x be real (so you are talking about real roots), and are a and b assumed to be positive? Possibly it may be true under some such conditions. But we can't help you prove it without (a) being sure of the real problem, and (b) seeing where you need help in your work. You put this under calculus; is there a reason to suppose that calculus methods will be of use? What methods have you learned that you think are applicable? You must log in or register to reply here. Share: Facebook X (Twitter) Reddit Pinterest Tumblr WhatsApp Email Share Link

• Forums
• Free Math Help
• Calculus

• This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register. By continuing to use this site, you are consenting to our use of cookies. Accept Learn more…

Top