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sat_chen

  • one year ago

did i get this partial integration question right?

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  1. sat_chen
    • one year ago
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    |dw:1359940501331:dw|

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

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

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

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

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

  7. hartnn
    • one year ago
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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...

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

  9. sat_chen
    • one year ago
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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)

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

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

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

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

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

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

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

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

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

  19. hartnn
    • one year ago
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    yes.

  20. sat_chen
    • one year ago
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    and really sorrmy internet is going berserk for some reason today college inter is acting funky for some reason

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

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

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

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

  25. hartnn
    • one year ago
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    lol.

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

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

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

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

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

  31. sat_chen
    • one year ago
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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

  32. hartnn
    • one year ago
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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

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

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

  35. hartnn
    • one year ago
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    u sure ?

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

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

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

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

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

  41. sat_chen
    • one year ago
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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

  42. sat_chen
    • one year ago
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    so can you tell me how you cancelled out the x in the B and the x in the A

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

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

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

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

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

  48. sat_chen
    • one year ago
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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

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

  50. sat_chen
    • one year ago
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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)

  51. sat_chen
    • one year ago
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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

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

  53. hartnn
    • one year ago
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    do what ?

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

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

  56. hartnn
    • one year ago
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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...

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

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

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