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Equations Quadratic in Form ;_;

  1. Dec 25, 2011 #1
    1. The problem statement, all variables and given/known data
    y^4 + 3y^2 - 4 = 0


    2. Relevant equations



    3. The attempt at a solution
    The base, I know, is y^2. So in order to make this a quadratic equation, we come up with an arbitrary variable, say α, which is equal to the base. Re-writing this, we get:

    (y^2)^2 + 3(y^2) - 4 = 0
    α = y^2
    α^2 + 3α - 4 = 0

    So now it's a quadratic equation.
    Factoring, we get:

    (α - 1) (α + 4) = 0

    Setting both of these equal to zero, we get:

    α = 1 and α = -4

    This next part is where I'm lost, so maybe one of you bright people can show me where I'm making my mistake.

    We start with α = -4.

    Since α = y^2, -4 must = y^2 too.

    -4 = y^2

    Solving this for y, we get:

    y = sqrt(-4) = 2i.

    However, in the text - without any explanation as to how this occurs (which is what is confusing me about these problems) - it says that the two solutions are 2i and -2i.

    Can someone please explain this step to me. D:
     
  2. jcsd
  3. Dec 25, 2011 #2
    Oh. Wait. I think I figured it out. When you take the square root of something we have to remember to put plus/minus, right? ;_; Maybe it's time to go back to radical rules! How embarrassing! Sorry, Physics Forums!
     
  4. Dec 25, 2011 #3
    Yep that's exactly what it is you can have to answers ±√.
     
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