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Ranges of simple quadratic functions

  1. Jun 19, 2012 #1
    Taking a general function with two quadratic eqs,
    Y = (ax^2+bx+c / px^2+qx+r)
    ax+bx+c = pyx^2+qxy+ry
    x^2(a-py) + x(b-qy)+(c-ry)=0
    The discriminant turns out to be a+ve or a 0.
    So,
    (b-qy)^2 - 4(a-py)(c-ry) >or= 0
    Now how does this equation yield a range for y?or can it?what are the next steps?
     
  2. jcsd
  3. Jun 19, 2012 #2

    HallsofIvy

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    You say you have "two quadratic equations" but I see three equations, none of which is quadratic.
     
  4. Jun 20, 2012 #3
    that is probably because of me using the m.physicsforum because of which i lack a "^2”(at the 2nd equation, the way you put it) and a couple of signs of implications after the first equation(which is the attempt towards the solution) and by saying i have two quadratic equations, i mean to say one quadratic eq over the other = quadratic fraction
     
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