What values satisfy this equation: ((2x-7)/(x3+3))<(9/x)<=x²-5x+9?

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Homework Help Overview

The discussion revolves around solving the compound inequality involving rational and polynomial expressions: ((2x-7)/(x3+3))<(9/x)<=x²-5x+9. The original poster expresses difficulty in determining the values of x that satisfy the second part of the inequality.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to analyze the inequality (9/x)<=x²-5x+9, with suggestions to manipulate the inequality by multiplying through by x². There is mention of using the rational root theorem for factorization.

Discussion Status

The discussion is ongoing, with some participants providing guidance on how to approach the problem. There is an acknowledgment of the need for the original poster to clarify their attempts at a solution.

Contextual Notes

Participants note the importance of adhering to forum guidelines regarding the presentation of attempts at a solution. There is also a reminder about the implications of multiplying by x², which is positive for all real non-zero x.

andreynr6
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Excuse me for my english;

Decide for all the x so this works out; ((2x-7)/(x3+3))<(9/x)<=x²-5x+9

where "<=" is the same as "and =".

Having some troubles getting the right values between (9/x)<=x²-5x+9, but otherwise it's fine..

hope you can help me!
 
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For future reference, please use the layout given to you when starting a thread in the homework help section, particularly the "attempt at a solution" part.

To find where x^2-5x+9\geq 9/x if you multiply through by x2 (a positive number for all real non-zero x) you will end up with a quartic that can be factorized easily using the rational root theorem. It should be easy from there where it's greater than zero.
 
Sorry about that Mentallic. I moved his thread to Homework Help from a general math forum, and asked him to post more of his work. I should have posted a note here as well.
 
Oh, no problem :smile:
 

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