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Unsure how to work algebra problem

  1. Nov 23, 2008 #1
    (s^2+4s+4)/(s^2+4s+5) = 1 - (1/(s^2+4s+5)



    3. The attempt at a solution

    I'm actually pretty far up into my upper level math classes (past calc 3), but I can't seem to figure out this one. How do you go from the left side to the right side of the equation?
     
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  3. Nov 24, 2008 #2

    symbolipoint

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    Are you trying to solve the equation? If you are, you do not "go from the left side to the right side". You should eliminate the rational expressions by multiplying both sides of the the equation by s^2 + 4s + 5.
     
  4. Nov 24, 2008 #3
    I'm not trying to solve the equation. It's for a circuits problem and in order to solve the problem, you have to set things up so you can do fraction decomposition. The right side is just what the solutions manual went to from what is on the left side. I just can't figure out how they got to the right side from the left side. Here is the original equation:

    F(s)= (s+2)^2/(s^2+4s+5).

    they factored out the numerator and then went to what I have on the right side. I just can't figure out how they got to it.
     
  5. Nov 24, 2008 #4

    Mentallic

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    This is really simple actually, and this idea will give you the boost you need:
    (note that the way I am expressing this is not official, I just want to give you the idea)

    [tex]\frac{a+m}{a+m+n}[/tex]

    [tex]= \frac{(a+m+n)-n}{a+m+n}[/tex]

    [tex]= 1-\frac{n}{a+m+n}[/tex]

    Notice how in the 2nd part there is an equal part of the numerator and the denominator, dealing with those is expressed as follows:

    [tex]\frac{ax+by}{x} = \frac{ax}{x}+\frac{by}{x} = a+\frac{by}{x}[/tex]
     
  6. Nov 24, 2008 #5

    symbolipoint

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    Are the numerator or denominator factorable? (I should try finding myself, but maybe your own effort might be just as helpful?).
    .......
    I thought about it a bit more: the numerator is factorable, but the denominator seems NOT factorable. (x __ 1) and (x __ 5), not useful.
    Are you allowed to use complex numbers?
     
  7. Nov 24, 2008 #6

    symbolipoint

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    Mentallic, I saw your post late. Your method seems unfamiliar based on "intermediate algebra" content; I'm not sure if I ever saw that type of process in "PreCalculus" either.
     
  8. Nov 24, 2008 #7
    Well, that is a really common trick, especially in integration. But, since the OP is higher in math classes, that should fit him/her perfectly fine.
     
  9. Nov 24, 2008 #8
    Yes, the denominator eventually is factored out with complex numbers (this is for a circuits class), but they get to what I have on the right side before doing that step.

    Mentallic, I'm not quite sure I follow what you posted for this particular problem.

    edit: I think it's throwing me off because the problem I'm dealing with has 3 terms on the top and bottom.
     
  10. Nov 24, 2008 #9

    Mentallic

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    I've never been 'taught' this process. I just happened to stumble upon it when trying to find the asymptotes for functions with the variable in the denominator.
    This process is quite simple to follow, I don't see why it shouldn't be in precalc or interm. algebra :smile:

    Anyway, thats how the text book would have jumped from the LHS to the RHS.
     
  11. Nov 24, 2008 #10
    What mentallic, is saying is this
    [tex]\frac{(s^2+4s+4)}{s^2+4s+5}=\frac{(s^2+4s+4+1)-1}{s^2+4s+5}=\frac{s^2+4s+5}{s^2+4s+5}-\frac{1}{s^2+4s+5}=1-\frac{1}{s^2+4s+5}[/tex]
     
  12. Nov 24, 2008 #11

    Mentallic

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    Hmm so you both aren't following what I posted?

    SnoggerT you were trying to prove that the LHS is equal to the RHS correct? But if you mean calculus in university, my help may be a little too primitive since I'm still in highschool.
     
  13. Nov 24, 2008 #12
    - Okay. I see now. I guess with it showing the n in the denominator and then moving it up to the numerator threw me off. Thanks.
     
  14. Nov 24, 2008 #13
    There is nothing wrong with Mentallic's method. It is short,nice, and easy!
     
  15. Nov 24, 2008 #14
    - I guess proving it would be right. I just wasn't sure how they got to the right hand side (proving one side to the other wasn't part of the problem). I don't remember things like this because I hadn't seen algebra in 10+ years when I started back in school for another degree.
     
  16. Nov 24, 2008 #15

    Mentallic

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    Yes exactly, but I try to avoid just giving the solution away so that I can avoid the authorities and their harsh punishments :cry:
     
  17. Nov 24, 2008 #16
    haha...I didn't really need the solution, I just needed to know what the 'n' really was in what you showed me. I was thinking way too much about it and overlooked that all I was doing was making the top and bottom equal for one part (to get the 1).
     
  18. Nov 24, 2008 #17

    Mentallic

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    I'm always willing to try explain something in more than 1 way :smile: Good luck with the rest of your problem!
     
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