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sat_chen

did i get this partial integration question right?

  • one year ago
  • one year ago

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  1. sat_chen
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    |dw:1359940501331:dw|

    • one year ago
  2. sat_chen
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    Answer |dw:1359940559335:dw|

    • one year ago
  3. sat_chen
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    can someone just check if that is the right answer

    • one year ago
  4. sat_chen
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    im still kind of confused about it

    • one year ago
  5. UnkleRhaukus
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    i think you've made a mistake somewhere,

    • one year ago
  6. sat_chen
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    ok i did it wrong apparently but i dont know how to do it still :( ?

    • one year ago
  7. 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...

    • one year ago
  8. sat_chen
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    ok i will type it out for you

    • one year ago
  9. 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)

    • one year ago
  10. sat_chen
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    then set the 2x^2 -12x+4=A/x+B/(x+2)+C/(x-2)

    • one year ago
  11. hartnn
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    here.... x^3-4x^2 = x^2 (x-4) and not x (x^2-4)

    • one year ago
  12. sat_chen
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    wow i cant believe i missed that

    • one year ago
  13. hartnn
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    since you know partial fractions , i think you'll get it in your next try..

    • one year ago
  14. sat_chen
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    sorry guys my intentet is acting weird

    • one year ago
  15. sat_chen
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    yea ill try the steps again and after im done pm you the steps

    • one year ago
  16. sat_chen
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    to see if i did it right thanks

    • one year ago
  17. hartnn
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    ok, sure. or you can post it here.. welcome ^_^

    • one year ago
  18. sat_chen
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    so would the partial thing be Ax+ B /x^2 +C/(x-4)

    • one year ago
  19. hartnn
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    yes.

    • one year ago
  20. 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

    • one year ago
  21. hartnn
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    A/x+ B /x^2 +C/(x-4)

    • one year ago
  22. sat_chen
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    wait isnt when the denominator is x^2 the top part is (Ax + B )/(x^2)

    • one year ago
  23. hartnn
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    its the same thing! (Ax+B)/x^2 = A/x + B/x^2

    • one year ago
  24. sat_chen
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    yep really had a long day today -.- after the superbowl drink and math dont go together

    • one year ago
  25. hartnn
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    lol.

    • one year ago
  26. sat_chen
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    ok after im done ill post the answer ty

    • one year ago
  27. sat_chen
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    so 2x^2-12x+4 = A/x + B/x^2 +c/(x-4)

    • one year ago
  28. sat_chen
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    and i solved for c which got me -3/16

    • one year ago
  29. sat_chen
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    |dw:1359956605483:dw|

    • one year ago
  30. sat_chen
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    and i plugged in as x = 4;

    • one year ago
  31. 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

    • one year ago
  32. 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

    • one year ago
  33. hartnn
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    what part of partial fractions, do you not get ?

    • one year ago
  34. sat_chen
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    if you put x = 0 you will get all 3 = 0

    • one year ago
  35. hartnn
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    u sure ?

    • one year ago
  36. sat_chen
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    |dw:1359957041709:dw| so you cant put 0

    • one year ago
  37. sat_chen
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    yea i got the c one though with is at x = 4

    • one year ago
  38. hartnn
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    ......+B (x-4)+.....

    • one year ago
  39. hartnn
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    A x (x-4) +B (x-4) +C (x^2)

    • one year ago
  40. sat_chen
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    |dw:1359957163440:dw|

    • one year ago
  41. 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

    • one year ago
  42. 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

    • one year ago
  43. hartnn
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    |dw:1359957466093:dw|

    • one year ago
  44. hartnn
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    |dw:1359957564998:dw|

    • one year ago
  45. sat_chen
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    shouldnt A/x be multiplied by |dw:1359957703525:dw|

    • one year ago
  46. hartnn
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    then denominator of A will become x^3 (x-4) which we don't want and is not common

    • one year ago
  47. sat_chen
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    hold on what i meant was isnt the format |dw:1359957835056:dw|

    • one year ago
  48. 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

    • one year ago
  49. hartnn
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    so, that i have SAME denominator for all the 3.

    • one year ago
  50. 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)

    • one year ago
  51. 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

    • one year ago
  52. hartnn
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    |dw:1359958265843:dw|its as simple as this...

    • one year ago
  53. hartnn
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    do what ?

    • one year ago
  54. sat_chen
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    oh i see i guess you could multiply what i did but it would cancel

    • one year ago
  55. sat_chen
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    i think i finally get it now

    • one year ago
  56. 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...

    • one year ago
  57. sat_chen
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    now if i had payed attention in my algebra classes this would be so much easier -.-

    • one year ago
  58. hartnn
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    no problem! because its easier for you now! right ?

    • one year ago
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