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**Help plz!**

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

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

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HallsofIvy

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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

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uart

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[tex] 4 z^3 - 3 z - 6 y = 0 [/tex]

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

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HallsofIvy

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

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I always assume calculator syntax if it's unclear.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))).

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