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Jadedry

  • one year ago

Resolve the following expression into its simplest partial fractions:

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  1. Jadedry
    • one year ago
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    the equation in question is: \[(x-2) / ( (x^2 +1)*(x-1)^2)\] I've managed to figure out that \[-0.5 / (x-1) ^2 \] is one of the partial fractions, but I'm having a lot of trouble with the others. Thanks in advance!

  2. BAdhi
    • one year ago
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    In this particular scenario you can decompose this to following partial fractions, \[\frac{x-2}{(x^2+1)(x-1)^2} = \frac{Ax+B}{x^2+1}+\frac{C}{(x-1)}+\frac{D}{(x-1)^2}\] So what you have to do now is to find the values of the A,B,C,D

  3. FireKat97
    • one year ago
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    @BAdhi can you please explain why its Ax + B and not simply A or B alone...

  4. Jadedry
    • one year ago
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    @Firecat97 Because \[(x^{2} + 1)\] is quadratic due to the fact that x is squared in the case of \[(x-1)^{2}\] , the x is not quadratic, but you have to come up with a numerator for \[(x-1) \] as well as \[(x-1)^{2}\]

  5. FireKat97
    • one year ago
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    ohhhhh okayyy thanks! we just started learning this stuff last lesson :)

  6. Jadedry
    • one year ago
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    @BAdhi I managed to get that far in my calculations, but I never come up with the right numerators? I think my method may be too complex, is there a simple way to find the numerator when a simple substitute for x will not cancel out all but one constant?

  7. FireKat97
    • one year ago
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    so D = -0.5 right?

  8. Jadedry
    • one year ago
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    @FireKat97 yes

  9. FireKat97
    • one year ago
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    okay thanks @Jadedry

  10. BAdhi
    • one year ago
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    There are mainly two ways to find the constant First one is the substituting values to the x and comparing numerator Second one is comparing the coefficients of each order of x The numerator is, \(x-2 = (Ax+B)(x^2-2x+1)+C(x^3-x^2+x-1)+D(x^2+1)\) now you have to compare the coefficients of some of the orders of x (choose the easiest one so that only one unknown or less number of coefficient is left) For example take \(x^3\) \(0 = A+C \implies C = -A\) from constant => \(-2 = B-C+D \implies B = -1.5+C\) compare another single such occation and youll find the value for C and thus values to others from above equations (i recommend coeff of \(x\))

  11. Jadedry
    • one year ago
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    Ah, I see you've expanded the equations I didn't do that. I don't understand how you got: 0= A +C I understand the second one, but I'm sorry to say the first one escapes me?

  12. BAdhi
    • one year ago
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    the coefficient of \(x^3\) on the left side is 0 the coefficient of the \(x^3\) on the right side comes only from \((Ax+\cdots)(x^2-\cdots) +C(x^3+\cdots)+\cdots = (A+C)x^3 +\cdots\) when equating the left side to right, 0 = (A+C)

  13. FireKat97
    • one year ago
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    would the value for C be 3/2?

  14. FireKat97
    • one year ago
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    no, 1/2?

  15. Jadedry
    • one year ago
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    @BAdhi Ah that is clearer, thank you. @FireKat97 No, I don't think those are the answers.

  16. FireKat97
    • one year ago
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    oh lol okay

  17. Jadedry
    • one year ago
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    @FireKat97 I think I came up with those numbers myself, but the textbook says they are wrong, no idea why.

  18. FireKat97
    • one year ago
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    hmm maybe C is -1

  19. BAdhi
    • one year ago
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    @FireKat97 If we take the numerator of (x-1)^2 as Cx+D jst like for (x^2+1) \[\frac{Cx+D}{(x-1)^2} = \frac{C(x-1)+(D+C)}{(x-1)^2}=\frac{C(x-1)}{(x-1)^2}+\frac{(C-D)}{(x-1)^2}= \frac{C}{(x-1)}+\frac{E}{(x-1)^2}\] So you can see it can furthur simplfied to that but you cant do it to x^2+1

  20. BAdhi
    • one year ago
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    there is a square at for the denominator of E its not visible :(

  21. Jadedry
    • one year ago
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    @FireKat97 Still not right, my book says the answer is 1??

  22. FireKat97
    • one year ago
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    soooo we have an E too now?

  23. FireKat97
    • one year ago
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    wait I think I see why C is 1 but I maybe wrong...

  24. FireKat97
    • one year ago
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    so do we get A -2B + C -1 = 0 then -C -2(-1.5 + C) + C -1 = 0 -C + 3 - 2C +C - 1 = 0 -2C = -2 C= 1? idkkkk

  25. Jadedry
    • one year ago
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    @FireKat97 actually that look pretty good

  26. FireKat97
    • one year ago
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    lololol legit?

  27. Jadedry
    • one year ago
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    @FireKat97 where did you get the eq ( A -2B + C -1 = 0 ) from though?

  28. FireKat97
    • one year ago
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    I did the same thing that @BAdhi was doing earlier, but I did it to the coefficients for x as he had suggested.

  29. BAdhi
    • one year ago
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    @FireKat97 Yes there is a E but also note that D has disappeard :P. so if you want you can replace E with D and you'd end up with the expression that i've used on the original question

  30. FireKat97
    • one year ago
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    ummmmmmmm soo the E is really just a D?

  31. FireKat97
    • one year ago
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    @Jadedry what does your textbook say the whole answer is btw?

  32. BAdhi
    • one year ago
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    @FireKat97 Yep

  33. Jadedry
    • one year ago
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    @FireKat97 it says the answer is: \[\frac{ 1}{ x-1 } - \frac{ 1 }{ 2(x-1)^2} - \frac{2x+1}{2(x^2+1)}\]

  34. Jadedry
    • one year ago
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    so, D is still -0.5 a= -1 b =- 0.5 c = 1 right @BAdhi ?

  35. BAdhi
    • one year ago
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    yes

  36. Jadedry
    • one year ago
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    ah i think I have it. So if you substitute 2 for x you get: 2a + b+ 5c +5d if you substitute c for a and -1.5 + c for b you have: -2c + c -1.5 +5c - 2.5 = 0 =4c - 4 = 0 - therefore c = 1 as the equation requires therefore, b= -1.5 +c b = -0.5 a = -c a = -1 and d remains -0.5 Yeah! got it!

  37. FireKat97
    • one year ago
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    Yup I think I got it too, thanks @Jadedry and @BAdhi :D

  38. Jadedry
    • one year ago
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    Thanks @BAdhi , never would have got it without that a = -c , you also gave pretty solid advice. Thanks @FireKat97 you managed to point me in the right direction Never would have done it without you guys, this problem has been bugging me for days.

  39. BAdhi
    • one year ago
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    ok cool.. u are welcome

  40. FireKat97
    • one year ago
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    :)

  41. Jadedry
    • one year ago
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    Okay, closing the board guys. Thanks again!

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