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anonymous

  • 5 years ago

can someone help me with partial fraction ? 4/x(x^2+4)

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  1. anonymous
    • 5 years ago
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    i might

  2. anonymous
    • 5 years ago
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    so the function is 4/( x * ( x^2 + 4) )?

  3. anonymous
    • 5 years ago
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    yes thats right

  4. anonymous
    • 5 years ago
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    yes thats right

  5. anonymous
    • 5 years ago
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    cool, so what's the first step? separate the function based upon the denominator, you have two functions mutliplied by eachother on the bottom: x * (x^2+4)

  6. anonymous
    • 5 years ago
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    so you could create a sum of two fractions: one with x as the denonm. and one with (x^2+4) as the denom. can you post those please?

  7. anonymous
    • 5 years ago
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    i don't quite get it.. ><

  8. anonymous
    • 5 years ago
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    np, ok, the goal is to express the function as the sum of other functions, which are actually "fractions"...where the denominators of these functions that you want to sum are the parts of the denominator of the larger function....clear?

  9. anonymous
    • 5 years ago
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    so basically, here you have 4 ?( x * (x^2 + 4) )...the larger function on the denom is (x * (x^2 + 4) )....so can you separate this into two functions?

  10. anonymous
    • 5 years ago
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    4 / x + 1/(x^2+4) im not very sure.. i suck at maths

  11. anonymous
    • 5 years ago
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    right now, ignore the numerator so you're right, but ignore 4...the setup is now: \[4/(x * (x^2 + 4) ) = A/(x) + B/(x^2+4)\]

  12. anonymous
    • 5 years ago
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    does that look familiar, like something in your text?

  13. anonymous
    • 5 years ago
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    but shouldn't it be A / x + Bx+c / (x^2+4)?

  14. anonymous
    • 5 years ago
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    ok, why do you think that you use "x" twice?

  15. anonymous
    • 5 years ago
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    because of the demo has a (x^2)?

  16. anonymous
    • 5 years ago
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    sorry, you are right about that equation not being quite right, because there is x^2+4...so you actually need to adjust the numerator of the fraction with (x^2+4)

  17. anonymous
    • 5 years ago
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    good catch! the equation becomes: \[4/(x∗(x2+4))=A/(x)+(Bx+C)/(x2+4) \]

  18. anonymous
    • 5 years ago
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    yeap.. that about it.. the only thing is i don't know how to solve for Bx+C

  19. anonymous
    • 5 years ago
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    ok, now do you think you can make these fractions so that they have a common denominator? hint: multiply by "1"

  20. anonymous
    • 5 years ago
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    i did a cross multipy which end up to be 4= A(x^2+4) + X(Bx+c)?

  21. anonymous
    • 5 years ago
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    ok, however you are adding these fractions...so you cannot cross multiply...you need to "give" the fractions the same denominator...and easy way would be multiply each fraction by 1, where 1 is actually a fraction with the numerator and denominator equal, in this case you would want to multiply A/x by (x^2+4)/(x^2+4)...see how this multiplying by 1 is OK?

  22. anonymous
    • 5 years ago
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    so the fraction on becomes: \[( A * (x^2 + 4))/(x * (x^2 +4))\]

  23. anonymous
    • 5 years ago
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    cool?

  24. anonymous
    • 5 years ago
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    ohh icic

  25. anonymous
    • 5 years ago
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    nice, now can you apply the concept to the other fraction? then post the new equation?

  26. anonymous
    • 5 years ago
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    A(x^2+4) / x(x^2+4) + Bx+C / x(x^2+4) am i right?

  27. anonymous
    • 5 years ago
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    almost, what about the numerator on the second fraction?

  28. anonymous
    • 5 years ago
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    the second fraction demo is x(x^2+4) isnt it same as the 1st fraction 's demo?

  29. anonymous
    • 5 years ago
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    yes, that's right, but the numerator is missing something...

  30. anonymous
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    \[(A * (x^2 + 4) ) /( x * x^2 +4 )) + ((Bx + C)*(????))/(x * (x^2 + 4))\]

  31. anonymous
    • 5 years ago
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    is it 1 ?

  32. anonymous
    • 5 years ago
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    why 1?

  33. anonymous
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    right now you've only multiplied the second fraction by (1/x)...so you didn't multiply by "1" like you did for the first with (x^2+4)/(X^2+4)

  34. anonymous
    • 5 years ago
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    then is it Bx+C * x(x^2+4)?

  35. anonymous
    • 5 years ago
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    so you're close, right?

  36. anonymous
    • 5 years ago
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    so i got to muplity Bx+C with X(x^2+4)?

  37. anonymous
    • 5 years ago
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    not quite, right it is multiplied by 1/x...see how you put x in the denominator? multiply the numerator by x too...so that you have multiplied by "1"..right?

  38. anonymous
    • 5 years ago
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    Bx+c * 1/x / x(x^2+4) ?? im confused

  39. anonymous
    • 5 years ago
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    no, you were very close with (Bx + c) /x * (x^2 + 4)...how did you get here from (Bx+c)/(x^2 + 4)?

  40. anonymous
    • 5 years ago
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    oh cancel out the x from the x(9x^2+4) but then the 1st fraction's demo isnt the same as 2nd one?

  41. anonymous
    • 5 years ago
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    typo error x(x^2+4)

  42. anonymous
    • 5 years ago
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    er, no canceling now

  43. anonymous
    • 5 years ago
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    actually do the opposite....so (Bx+c)/(x^2+4) becomes ((Bx+c)*x)/((x^2+4)*x)...see how you multiplied numerator and denominator by x because you mutliplied the fraction by "1" where "1" = x/x

  44. anonymous
    • 5 years ago
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    so, what are the two fractions now, now that they have common denominators of x*(x^2+4)...rewrite the equation

  45. anonymous
    • 5 years ago
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    A( x^2+4) / x(x^2+4) + Bx+c * (x) / x (x^2+4)

  46. anonymous
    • 5 years ago
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    right on

  47. anonymous
    • 5 years ago
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    so the entire equation looks like this: \[4/(x * (x^2 + 4)) = ((A * (x^2 + 4)) + ((Bx + c) *(x)))/(x * (x^2 + 4))\]

  48. anonymous
    • 5 years ago
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    yeap..

  49. anonymous
    • 5 years ago
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    nice, now you can ignore denominators for the time being...so we actually want to look at the equality between the numerators

  50. anonymous
    • 5 years ago
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    \[4 = (A *(x^2 + 4)) + (Bx+c)*x \]

  51. anonymous
    • 5 years ago
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    this eqn isnt it same as what just now i had cross multiply them together?

  52. anonymous
    • 5 years ago
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    i'm not sure what you mean by cross multiply...

  53. anonymous
    • 5 years ago
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    as stated just now A/x + Bx+c/ x^2+4 i cross mutiply them.. meaning. A * (x^2+4) and x* (bx+c)

  54. anonymous
    • 5 years ago
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    in the original equation you're saying that you multiplied everything by the denominator?

  55. anonymous
    • 5 years ago
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    yea.

  56. anonymous
    • 5 years ago
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    cross multiply refers to when you have two fractions: like a/b = c/d , cross multiplying would yield a*d = c*b...

  57. anonymous
    • 5 years ago
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    that is what cross-multiplying technically is...it is best to the procedure as above so as to not move too hastily and miss a step...

  58. anonymous
    • 5 years ago
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    ohh..

  59. anonymous
    • 5 years ago
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    so anyway, we are close to solving

  60. anonymous
    • 5 years ago
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    4=(A∗(x2+4))+(Bx+c)∗x

  61. anonymous
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    \[4 = (A * (x^2 + 4)) +(Bx + c)*(x)\] expand this

  62. anonymous
    • 5 years ago
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    4 = Ax^2 + 4A + Bx^2 + cx?

  63. anonymous
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    \[4 = (A * (x^2 + 4)) + (Bx + C) * x = Ax^2 + 4A + Bx^2 + Cx\]

  64. anonymous
    • 5 years ago
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    so, now, let's consider this, both equations are assumed to be equal for all values of x...right, because it's an equation

  65. anonymous
    • 5 years ago
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    so you see how only one term does not have x in it?

  66. anonymous
    • 5 years ago
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    the 4A

  67. anonymous
    • 5 years ago
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    right on! so now we need to figure out what each letter should equal

  68. anonymous
    • 5 years ago
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    therefore A = 1

  69. anonymous
    • 5 years ago
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    dude, nice!

  70. anonymous
    • 5 years ago
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    yeah, the simplest case was to consider that x = 0, now we have that A =1...so let's find C

  71. anonymous
    • 5 years ago
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    there are no x terms on the left side, so, what do you think C equals?

  72. anonymous
    • 5 years ago
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    but as from just now eqn 4 = A(x^2+4) + x(Bx+c) can i sub X = 0 for Bx+c(x) so that 4=A(0^2+4)? which turn out to be 4 = 4A?

  73. anonymous
    • 5 years ago
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    Bx^2+Cx...that is the idea

  74. anonymous
    • 5 years ago
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    Ax^2+4A+Bx^2+Cx, ok we should group these with like terms --> (A+B)*x^2 +(C)*x + 4A

  75. anonymous
    • 5 years ago
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    (A+B)*x^2 +(C)*x + 4A = 4

  76. anonymous
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    you figured out that A = 1 by looking at the case when x=0

  77. anonymous
    • 5 years ago
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    yea

  78. anonymous
    • 5 years ago
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    and that where i struck finding b and c

  79. anonymous
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    now, on the right side here there are no terms with x^2 nor x...so what do you think C would be equal to?

  80. anonymous
    • 5 years ago
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    C = 0?

  81. anonymous
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    right on!

  82. anonymous
    • 5 years ago
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    now, you know A=1, C=0, now...to find B....so we look at where we have B, only ing (A+B)*x^2

  83. anonymous
    • 5 years ago
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    which (1 + B) *x^2 because A=1

  84. anonymous
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    again, there are no x^2 terms on the other side..so B must equal what?

  85. anonymous
    • 5 years ago
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    3?

  86. anonymous
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  87. anonymous
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    why 3?

  88. anonymous
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    the idea is that on the other side the term is 4 + 0*x^2, so we have (1+B)*x^2 = (0)*x^2

  89. anonymous
    • 5 years ago
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    ohh.. its -1

  90. anonymous
    • 5 years ago
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    yes, so what are A, B and C

  91. anonymous
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    A= 1 B= -1 C= 0

  92. anonymous
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    one you have them all, substitute them back in the fractions

  93. anonymous
    • 5 years ago
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    oicic

  94. anonymous
    • 5 years ago
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    where they were separted

  95. anonymous
    • 5 years ago
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    in here: 4/(x*(x^2+4))=A/(x)+B/(x^2+4)

  96. anonymous
    • 5 years ago
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    ok got it.. thx a lot man~ sry to wate such a long time.. hope u don't mind

  97. anonymous
    • 5 years ago
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    no problem, now do you think you can solve another partial fraction problem?

  98. anonymous
    • 5 years ago
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    not really.. cos' theres a lot of type of partial fraction..

  99. anonymous
    • 5 years ago
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    well the premise is the same....separate the denominator into a product of functions...then create the letter fractions with these smaller functions...solve for the letters...

  100. anonymous
    • 5 years ago
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    just use this conversation as notes....try another and good luck!

  101. anonymous
    • 5 years ago
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    ok thanks a lot

  102. anonymous
    • 5 years ago
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    np

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