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satellite73Best ResponseYou've already chosen the best response.3
method for finding \(b\) in \[\frac{5x^2+20x+6}{x(x+1)^2}=\frac{a}{x}+\frac{b}{x+1}+\frac{c}{(x+1)^2}\]
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
in terms of a and c and x ?
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
@mahmit2012 method with derivatives
 one year ago

LibnizBest ResponseYou've already chosen the best response.0
what's wrong with ordinary method of finding partial differential?
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
@mahmit2012 had a snap method, but i don't understand it
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
just a question: do wolfram has solution?
 one year ago

alienbrainBest ResponseYou've already chosen the best response.0
they have something ugly...
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
we 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
 one year ago

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

alienbrainBest ResponseYou've already chosen the best response.0
oh the original question was an integral, btw.
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
something to do with a derivative but i don't see it
 one year ago

alienbrainBest ResponseYou've already chosen the best response.0
what's your method for B
 one year ago

LibnizBest ResponseYou've already chosen the best response.0
is there link to where it was done in snappier way?
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
looks 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
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
but is sure is snapp!
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
I think we can solve this making three equations....
 one year ago

alienbrainBest ResponseYou've already chosen the best response.0
i meant did you have your own way, satellite?
 one year ago

alienbrainBest ResponseYou've already chosen the best response.0
a slower... sane approach
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.3
ooooooooooooooh!!!! multiply both sides by \((x+1)^2\)!!!!!
 one year ago

alienbrainBest ResponseYou've already chosen the best response.0
but that's not right i guess
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
how cool is that? i love it
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
@alienbrain yes i wrote my method before solved \(6+B=5\) but it required some visualization
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
dw:1344137128472:dw
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
it is already solved, i was trying to understand @mahmit2012 approach
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
now 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 \(B=1\)
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.3
dw:1344137486006:dw
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
@mahmit2012 my less sophisticated method was to visualize the coefficient of \(x^2\) as \(6+B\)
 one year ago

sauravshakyaBest 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
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.3
dw:1344137568037:dw
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
yes i see thanks i didn't understand that you multiplied on both sides
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.3
dw:1344137675922:dw
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
i think i would not want to use this method if i had to take the derivative repeatedly
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.3
this is the shortest method to get cofficiant in partial fraction problem >
 one year ago

alienbrainBest ResponseYou've already chosen the best response.0
back, open study lagged me out
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
yeah it was getting laggy hope we have worn this topic out sufficiently we have the answer and a plethora of methods
 one year ago

alienbrainBest ResponseYou've already chosen the best response.0
so 5=6 + B are you just combining like terms there
 one year ago

alienbrainBest ResponseYou've already chosen the best response.0
good thing you can't throw a math book at me:)
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
ok lets go back back back
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
first term on the left is \(A(x+1)^2=6(x+1)^2\) and we also have a term that looks like \(Bx(x+1)\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
whatever 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 \(Bx^2\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
on the other hand we know we have to end up with \(5x^2\) that tells you \(6+B=5\) and so \(B=1\)
 one year ago

alienbrainBest ResponseYou've already chosen the best response.0
ok, i got it... but the method seems to vary for each letter
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.3
ok 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
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
if you have no repeated factors, the it always works
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
well, if they are linear
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
it 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
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
ok 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
 one year ago

alienbrainBest ResponseYou've already chosen the best response.0
eh.. integral of 6/x ?
 one year ago
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