Algebra Attachment: Solve the Problem Now

  • Thread starter Thread starter viren_t2005
  • Start date Start date
  • Tags Tags
    Algebra
AI Thread Summary
The discussion focuses on solving a mathematical problem attached in the thread, which is identical to a previously posted question in general maths. The primary task is to solve for "a" to ensure the expression equals zero, followed by a similar solution for "b." Participants emphasize the simplicity of the algebra involved. The conversation highlights the importance of addressing the problem directly without unnecessary downloads. Clear solutions for both variables are sought to resolve the issue effectively.
viren_t2005
Messages
20
Reaction score
0
The question has been attached .
 

Attachments

Physics news on Phys.org
don't download the link, it is the identical question to that posted under the same title in general maths by sarvesh and that has been moved to, presumably, precalc homeworlk.
 
Simply solve for "a" so that the expression is always zero. Simple algebra there.

Do the same for "b", and there you go.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top