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Help !

  1. Nov 30, 2003 #1
    Help plz!

    How can I change this y=2/3x^(3/2)-1/2x^(1/2) in term of y, as x=...???
  2. jcsd
  3. Nov 30, 2003 #2
    I hate fractions, so multiply through by 6. Factor out x1/2 on the right. Square each side. Expand the squared terms, distribute the x, subtract 36y2 from each side and apply the cubic formula (I don't know what it is but I know it exists).
  4. Dec 1, 2003 #3


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    StephenPrivitera was assuming you meant

    y= (2/3)x^(3/2)-(1/2)x^(1/2).

    What you wrote could as easily be interpreted as

    y= 2/(3x^(3/2)-1/(2x^(1/2))).

    Please use parentheses to make your meaning clear.

    In a bit more detail, what he said was: multiply the equation by 6 to get
    6y= 4x^(3/2)- 3x^(1/2)= (4x- 3)x^(1/2)
    36y^2= (4x-3)^2(x)= (16x^2- 24x+ 9)x
    = 16x^3- 24x^2+ 9x

    Which you can write as 16x^3- 24x^2+ 9x- 36y^2= 0 and solve as a cubic equation.

    (You can see I have absolutely nothing to do this morning. Well, nothing I want to do!)
  5. Dec 1, 2003 #4


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    The easiest approach is not to square each side but instead to subsitute [tex]z = \sqrt{x}[/tex] which leads directly to the reduced cubic of,

    [tex] 4 z^3 - 3 z - 6 y = 0 [/tex]
  6. Dec 1, 2003 #5
    I have to solve the equation 2/3x^(3/2)-1/2x^(1/2) for "X", as x=....
  7. Dec 1, 2003 #6


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    Yes, that was what you said before and you got three replies telling you how to do that.
  8. Dec 1, 2003 #7
    I always assume calculator syntax if it's unclear.
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