sat_chen
did i get this partial integration question right?
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sat_chen
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|dw:1359940501331:dw|
sat_chen
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|dw:1359940559335:dw|
sat_chen
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can someone just check if that is the right answer
sat_chen
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im still kind of confused about it
UnkleRhaukus
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i think you've made a mistake somewhere,
sat_chen
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ok i did it wrong apparently but i dont know how to do it still :( ?
hartnn
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so, denominator = x^2 (x-4)
do you know partial fractions method ?
also, if you show your work, we can quickly spot the error, and continue from that step...
sat_chen
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ok i will type it out for you
sat_chen
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so first what i did was split the denominator to x(x^2-4)
then split that to (x+2)(x-2)
sat_chen
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then set the 2x^2 -12x+4=A/x+B/(x+2)+C/(x-2)
hartnn
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here....
x^3-4x^2 = x^2 (x-4)
and not x (x^2-4)
sat_chen
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wow i cant believe i missed that
hartnn
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since you know partial fractions , i think you'll get it in your next try..
sat_chen
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sorry guys my intentet is acting weird
sat_chen
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yea ill try the steps again and after im done pm you the steps
sat_chen
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to see if i did it right thanks
hartnn
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ok, sure.
or you can post it here..
welcome ^_^
sat_chen
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so would the partial thing be Ax+ B /x^2 +C/(x-4)
hartnn
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yes.
sat_chen
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and really sorrmy internet is going berserk for some reason today college inter is acting funky for some reason
hartnn
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A/x+ B /x^2 +C/(x-4)
sat_chen
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wait isnt when the denominator is x^2 the top part is (Ax + B )/(x^2)
hartnn
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its the same thing!
(Ax+B)/x^2 = A/x + B/x^2
sat_chen
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yep really had a long day today -.- after the superbowl drink and math dont go together
hartnn
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lol.
sat_chen
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ok after im done ill post the answer ty
sat_chen
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so 2x^2-12x+4 = A/x + B/x^2 +c/(x-4)
sat_chen
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and i solved for c which got me -3/16
sat_chen
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|dw:1359956605483:dw|
sat_chen
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and i plugged in as x = 4;
sat_chen
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to get c = -3/16 but you cant solve for the other A and B is there another method to do this? i dont get partial fractions
hartnn
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to find c, put x=4
to find b, put x=0
to find a, put x=any other value, say x=1
hartnn
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what part of partial fractions, do you not get ?
sat_chen
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if you put x = 0 you will get all 3 = 0
hartnn
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u sure ?
sat_chen
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|dw:1359957041709:dw| so you cant put 0
sat_chen
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yea i got the c one though with is at x = 4
hartnn
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......+B (x-4)+.....
hartnn
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A x (x-4) +B (x-4) +C (x^2)
sat_chen
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|dw:1359957163440:dw|
sat_chen
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wait can you tell me how you got that i got the trigonometric integral stuff but this stuff is confusing me
sat_chen
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so can you tell me how you cancelled out the x in the B and the x in the A
hartnn
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|dw:1359957466093:dw|
hartnn
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|dw:1359957564998:dw|
sat_chen
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shouldnt A/x be multiplied by |dw:1359957703525:dw|
hartnn
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then denominator of A will become x^3 (x-4)
which we don't want and is not common
sat_chen
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hold on what i meant was isnt the format |dw:1359957835056:dw|
sat_chen
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|dw:1359958002008:dw|
i guess what i mean is in this for example why did you only pick out x(x-4) for A only (x-4) for b and x^2 for C im really confused about that
hartnn
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so, that i have SAME denominator for all the 3.
sat_chen
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and also because in previous problems i did it was like 4x^2 + 54 x + 134 = A/(x-1) +B/(x+5) +C/(x+3)
which i got to 4x^2 + 54x + 134 = A(x+5)(x+3) + B(x-1)(x+3)+C(x-1)(x+5)
sat_chen
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do you have to do this when the denominator has a square on the x? or is it dependent on the problem
hartnn
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|dw:1359958265843:dw|its as simple as this...
hartnn
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do what ?
sat_chen
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oh i see i guess you could multiply what i did but it would cancel
sat_chen
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i think i finally get it now
hartnn
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in previous problem, to make the common denominator as (x-5)(x+5)(x+3)
so, you multiplied and divided by (x+5)(x+3) for A
yes, you can say that also...
sat_chen
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now if i had payed attention in my algebra classes this would be so much easier -.-
hartnn
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no problem! because its easier for you now! right ?