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Math Help - Radical Equations #3

  1. #1
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    Post Radical Equations #3

    solve

    My attempt:





    when I put it into the quadratic formula:

    the answer is suppose to be j = -2/3 what did i do wrong?
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  2. #2
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    Re: Radical Equations #3

    Are you sure you copied this down right? I'm getting a negative radical when I use the quadratic equation.

    Right away, plugging in \frac{-2}{3} makes \sqrt{j+\frac{1}{3}}=\sqrt\frac{-1}{3}
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  3. #3
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    Re: Radical Equations #3

    yes,

    I wrote it down correctly, can you comment on my attempt?
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  4. #4
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    Re: Radical Equations #3

    Quote Originally Posted by Mathnood768 View Post
    solve
    My attempt:


    Make it 3j+1=12j^2+12j+3. It is easier to work with.
    Last edited by Plato; June 21st 2013 at 11:55 AM.
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  5. #5
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    Re: Radical Equations #3

    Hello, Mathnood768!

    I wrote it down correctly. . No, I don't think so.

    If the answer is  j = \text{-}\tfrac{2}{3}
    . . the equation could be: . \sqrt{\frac{j+1}{3}} + 5j \:=\: 3i - 1
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