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