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Math Help - Substitution to transform to quadratic form

  1. #1
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    Substitution to transform to quadratic form

    Can anyone help me with this?

    3=\frac{1}{(x+1)^2} + \frac{2}{(x+1)}

    I sub u={(x+1)^-1}

    then flip them and rearrange so I get u^2+\frac{u}{2}-3=0

    then mutiply all by two to get rid of fraction.

    then I factor and end up with (2u-3)(u+2)=0

    I think the problems begin when I sub back in the {(x+1)^-1}... doing something wrong and not getting book's solutions.





    btw those -1's are supposed to be superscripted, not sure why its not working :/
    Last edited by hexrei; November 6th 2012 at 05:59 PM.
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  2. #2
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    Re: Substitution to transform to quadratic form

    if u = \frac{1}{x+1} , then the correct substitution would be ...

    3 = u^2 + 2u

    try again ...
    Thanks from MarkFL, hexrei and Petrus
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  3. #3
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    Re: Substitution to transform to quadratic form

    \frac{1}{a}+ \frac{1}{b} "flipped" does NOT give "a+ b".
    Thanks from MarkFL and hexrei
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    Re: Substitution to transform to quadratic form

    Quote Originally Posted by skeeter View Post
    if u = \frac{1}{x+1} , then the correct substitution would be ...

    3 = u^2 + 2u

    try again ...
    I appreciate the help and of course I'm going to try again. Why would I have taken the time to post it otherwise
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    Re: Substitution to transform to quadratic form

    OK, so I have wrapped my head around that early error I made.

    I'm getting tripped up after the factorization still though, and the book unfortunately has example subbing with negative exponents with single variable but none with a constant and a variable plus negative exponent.

    So after I sub x+1^{-1} back in. I will only list one of the zeros since I think once I see what I am missing the second shouldn't be hard.

    my factorization is now (u+3)(u-1)=0

    which after subbing back in the first one yields me

    \frac{1}{x+1} + 3

    from there I tried finding a common denominator

    \frac{1}{x+1} + \frac{3x+3}{x+1}

    and adding them like so

    \frac{3x+4}{x+1}

    But not sure if that was correct, or what to do next. Can someone give me a clue please?
    Last edited by hexrei; November 6th 2012 at 06:17 PM.
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  6. #6
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    Re: Substitution to transform to quadratic form

    When is a rational expression equal to zero?
    Thanks from hexrei and Petrus
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    Re: Substitution to transform to quadratic form

    Quote Originally Posted by MarkFL2 View Post
    When is a rational expression equal to zero?
    oooh..... when the numerator is zero? So x would be -4/3. Got it. Thanks
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