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satellite73
 4 years ago
partial fraction madness, continuation
satellite73
 4 years ago
partial fraction madness, continuation

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0method for finding \(b\) in \[\frac{5x^2+20x+6}{x(x+1)^2}=\frac{a}{x}+\frac{b}{x+1}+\frac{c}{(x+1)^2}\]

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.0in terms of a and c and x ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@mahmit2012 method with derivatives

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what's wrong with ordinary method of finding partial differential?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@mahmit2012 had a snap method, but i don't understand it

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.0just a question: do wolfram has solution?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0they have something ugly...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we get \(a=\frac{6}{1}=6\), and \(c=\frac{9}{1}=9\) but \[b=5+0\frac{6}{(1)^2}=1\] step i don't get

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=integrate+%285x%5E2%2B20x%2B6%29+%2F+%28x%5E3%2B2x%5E2%2Bx%29

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh the original question was an integral, btw.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0something to do with a derivative but i don't see it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what's your method for B

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is there link to where it was done in snappier way?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0looks like the method was, take \[5x^2+20x+6\] divide by \(x\) get \[5x+20+\frac{x}{6}\] take the derivative get \[5\frac{6}{x^2}\] and then evaluate at \(x=1\) why this works is going to bug me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but is sure is snapp!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think we can solve this making three equations....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i meant did you have your own way, satellite?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a slower... sane approach

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i plugged in A and C.. expanded it all out... plugged in 1 for x and got B=3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ooooooooooooooh!!!! multiply both sides by \((x+1)^2\)!!!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but that's not right i guess

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how cool is that? i love it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@alienbrain yes i wrote my method before solved \(6+B=5\) but it required some visualization

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1344137128472:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is already solved, i was trying to understand @mahmit2012 approach

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now i get it, maybe i can remember is @alienbrain do not multiply out, you will make an algebra mistake, it is a pain we know \(6+B=5\) and so \(B=1\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1344137486006:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@mahmit2012 my less sophisticated method was to visualize the coefficient of \(x^2\) as \(6+B\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@satellite73 are u sure your answer is correct.... because I got a=15 b=10 and c=1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1344137568037:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes i see thanks i didn't understand that you multiplied on both sides

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1344137675922:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think i would not want to use this method if i had to take the derivative repeatedly

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is the shortest method to get cofficiant in partial fraction problem >

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0back, open study lagged me out

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah it was getting laggy hope we have worn this topic out sufficiently we have the answer and a plethora of methods

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 5=6 + B are you just combining like terms there

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0good thing you can't throw a math book at me:)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok lets go back back back

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0first term on the left is \(A(x+1)^2=6(x+1)^2\) and we also have a term that looks like \(Bx(x+1)\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0whatever all the other mess is when you multiply out, the only way to get \(x^2\) is from the first term, which will give you \(6x^2\) and from the second term which will give you \(Bx^2\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0on the other hand we know we have to end up with \(5x^2\) that tells you \(6+B=5\) and so \(B=1\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, i got it... but the method seems to vary for each letter

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i was hoping to avoid critical thinking and just be able to do it by a process :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok you can use one method if you like you can multiply this mess out and solve a 3 by 3 system, but i try to avoid that if at all possible because it is a ton of work try to look for values of \(x\) that will make everything zero obviously that will not work all the time when you have repeated factors but i find it much easier to use that method as far as i can once you have some of the constants, the others will be easier to find, even if you have a system of equations

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you have no repeated factors, the it always works

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well, if they are linear

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is also very easy to make an algebra mistake when multiplying out and combining like terms, much few mistakes plugging in numbers and evaluating in any case if you were going to use "one method" you would have ended up with the equation \(A+B=5\) but think of how much easier if you know \(A=6\) already and we got \(A=6\) with almost no work

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok enough partial fractions for one night i wish @eliassaab was here to show another method it has to do with taking the limit as \(x\to \infty\) which for a rational function you can pretty much eyeball, and then determining what a constant is from that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0eh.. integral of 6/x ?
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