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How would you solve this for y?

  1. Jul 11, 2011 #1
    x=-6y^2 + 4y
     
  2. jcsd
  3. Jul 11, 2011 #2

    mathman

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    Gold Member

    Use the quadratic formula.
     
  4. Jul 11, 2011 #3
    Try completing the square (i.e. as-if it was just a normal quadratic equation of a single variable)... it should work.
     
  5. Jul 11, 2011 #4
    okay, so plotting BOTH of these graphs will give me a perpendicular graph of y=-6x^2+4x, correct? each of the functions from the quadratic formula should be each half of a perpendicular parabola.
     
  6. Jul 11, 2011 #5
    hhhmmm...I just took the equation and put it like this:

    -6y2 + 4y - x = 0

    and solved it using the quadratic formula, so

    y = ( -b + sqrt(b2 - 4ac) ) / 2a

    where

    a=-6
    b=4
    c= -x

    and the other solution to y is with the negative value of the square root.

    I get something like this:

    y = 1/3 + sqrt( (2-3x)/18 )

    and, of course,

    y = 1/3 - sqrt( (2-3x)/18 )

    which tells me that x cannot be larger than 2/3, otherwise the radical becomes negative.

    So, you can plot y versus x for -inf <= x <= 2/3
     
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