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El_Arrow

  • one year ago

i need help with partial fractions problem

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  1. El_Arrow
    • one year ago
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    \[\frac{ 2x }{ (x-1)^3 }\]

  2. El_Arrow
    • one year ago
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    @zepdrix @ganeshie8

  3. zepdrix
    • one year ago
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    You have a repeated linear factor in the denominator: \(\large\rm (x-1)\) is linear. So do you remember how to break that down? I'll give an example:\[\large\rm \frac{4x+2}{(x+1)^2}=\frac{A}{x+1}+\frac{B}{(x+1)^2}\]And then you would solve for A and B.

  4. zepdrix
    • one year ago
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    \[\large\rm \frac{2x}{(x-1)^3}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}\]Understand the initial setup? :O

  5. El_Arrow
    • one year ago
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    yeah

  6. El_Arrow
    • one year ago
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    i used x=1 and got 2 for c is it correct?

  7. zepdrix
    • one year ago
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    mmm yah that sounds right! :)

  8. El_Arrow
    • one year ago
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    getting the other 2 looks complicated

  9. zepdrix
    • one year ago
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    Hmm I think I see a neat trick for the other ones :) Obviously we'd like to avoid expanding everything right?

  10. El_Arrow
    • one year ago
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    yeah

  11. El_Arrow
    • one year ago
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    okay let me try

  12. zepdrix
    • one year ago
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    Oh I made a boo boo.. my bad :( \(\large\rm A\ne B\) I forgot about the C=2 part on the end, woooops.. back up :O

  13. zepdrix
    • one year ago
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    Back to the drawing board >.< woops woops

  14. zepdrix
    • one year ago
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    I still like the idea of plugging in x=0, and x=-1 it will give you a system of 2 equations with A and B.

  15. zepdrix
    • one year ago
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    It's probably about the same amount of work as expanding though

  16. El_Arrow
    • one year ago
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    well i have the final answer at the back of the book and its |dw:1442723235969:dw|

  17. El_Arrow
    • one year ago
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    so c is not equal to 2

  18. zepdrix
    • one year ago
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    Really? Hmm wolfram is saying that we should have c=2 and b=2 https://www.wolframalpha.com/input/?i=partial+fraction+decomposition+%282x%29%2F%28x-1%29%5E3

  19. zepdrix
    • one year ago
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    maybe typo in the book :O

  20. El_Arrow
    • one year ago
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    i guess there is a typo in the book

  21. El_Arrow
    • one year ago
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    so how does it get B=2?

  22. zepdrix
    • one year ago
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    I keep running into mistakes when I try to plug values in. I'm just gonna expand it out and see if we can get the right values...

  23. El_Arrow
    • one year ago
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    okay im gonna try that too

  24. zepdrix
    • one year ago
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    \[\large\rm 2x=A(x-1)^2+B(x-1)+C\]\[\large\rm 2x=A(x-1)^2+B(x-1)+2\]\[\large\rm 2x=A(x^2-2x+1)+B(x-1)+2\]\[\large\rm 2x=Ax^2-2Ax+A+Bx-B+2\]Then, matching up the squares,\[\large\rm 0x^2=Ax^2\qquad\implies\qquad 0=A\]Matching up the first degree terms,\[\large\rm 2x=-2Ax+Bx\qquad\implies\qquad 2=-2A+B\]And matching up the constant terms,\[\large\rm 0=A-B+2\]

  25. El_Arrow
    • one year ago
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    so |dw:1442723741155:dw|

  26. El_Arrow
    • one year ago
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    thanks for helping me figuring this one out

  27. zepdrix
    • one year ago
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    Ya I guess we shoulda just expanded from the beginning D: wasn't as tough as i thought it would be

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