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Quadratic equation troubles

  1. Jan 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve 4y-y2=x for y.

    3. The attempt at a solution

    First I tried using the quadratic equation with a=-1 b=4 c=-x

    y=(-b±(42-4(-1)(-x))(1/2))/2(-1)

    That got me this far: y=2±(8-2x)(1/2)

    Then I checked wolfram and now I am confused
    y=±2-sqrt(4-x
    . Did I not approach this corectly?
     
  2. jcsd
  3. Jan 18, 2013 #2

    vela

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    Your approach is fine. You just made an algebra mistake while simplifying.

    You had ##y = \frac{-4 \pm \sqrt{16 - 4x}}{-2}##. Note that you can't simply divide the stuff inside the radical by 2 when you bring the 2 from the denominator inside.
     
  4. Jan 20, 2013 #3
    Thank you for the quick reply and sorry for my slow thanks. How very typical of me that it's some algebraic error :blushing: Thanks again!
     
  5. Jan 20, 2013 #4

    SammyS

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    As the originator of the thread, you don't really need to use a Spoiler.
     
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