Help plz!

[gigi9]

How can I change this y=2/3x^(3/2)-1/2x^(1/2) in term of y, as x=...???

[StephenPrivitera]

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).

[HallsofIvy]

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)
so
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!)

[uart]

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

4 z^3 - 3 z - 6 y = 0

[gigi9]

I have to solve the equation 2/3x^(3/2)-1/2x^(1/2) for "X", as x=....

[HallsofIvy]

Yes, that was what you said before and you got three replies telling you how to do that.

[StephenPrivitera]

Originally posted by HallsofIvy
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))).

I always assume calculator syntax if it's unclear.