Problem simplifying an equation

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The discussion focuses on the steps needed to simplify the equation P = a²P + b²P²/(P+1) into the quadratic form P² - (a² + b² - 1)P - b² = 0. Initial steps include rearranging the equation by moving a²P to the left and multiplying both sides by (P + 1). Despite these manipulations, the user expresses confusion about reaching the final solution, suggesting a potential error in the transformation. They highlight that substituting specific values leads to inconsistencies, indicating that the simplification process may not be valid. The conversation underscores the challenges of solving complex equations in algebra.
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Homework Statement


What is the steps between the two equations?


Homework Equations


P = a^{2}P+b^{2}P^{2}/(P+1) \Rightarrow P^{2}-(a^{2}+b^{2}-1)P-b^{2} = 0

Thanks!
 
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mr.t said:
What is the steps between the two equations?

P = a^{2}P+b^{2}P^{2}/(P+1) \Rightarrow P^{2}-(a^{2}+b^{2}-1)P-b^{2} = 0

Thanks!

Hi mr.t! :smile:

i] put the a2P on the left

ii] multiply both sides by (P + 1) :smile:
 
Hello Tiny-tim!
Putting the a^{2}P to the left and multiply both sides with (P+1) gives:

(P+1)(P-a^{2}P) = b^{2}P^{2} \Rightarrow
\Rightarrow P^{2}-a^{2}P^{2}+P-a^{2}P-b^{2}P^{2} = 0 \Rightarrow
\Rightarrow P^{2}(1-a^{2}-b^{2})+P(1-a^{2}) = 0

But I still can't come to the final solution? :S
 
I don't think the two steps are equivalent. Let b=0, then a^2=1 if P \neq 0. Substituting these into the second equation, we have a^2 = 1 only if P = 0
 
It could be an error in the solution then no fun try to learn stuff when you end up spending your time figuring out the imposible...

thanks for your time anyway!
 
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

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