How can I change this y=2/3x^(3/2)-1/2x^(1/2)

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Discussion Overview

The discussion revolves around the manipulation of the equation y=2/3x^(3/2)-1/2x^(1/2) to express x in terms of y. Participants explore various methods for solving this equation, including algebraic manipulation and substitution techniques.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests multiplying through by 6, factoring out x^(1/2), squaring both sides, and applying the cubic formula, although they admit uncertainty about the cubic formula itself.
  • Another participant clarifies the interpretation of the original equation, emphasizing the importance of parentheses for clarity and providing a detailed algebraic manipulation leading to a cubic equation.
  • A different approach is proposed by substituting z = √x, which simplifies the equation to a reduced cubic form, 4z^3 - 3z - 6y = 0.
  • One participant expresses frustration over the use of fractions and suggests an alternative method to avoid squaring both sides.
  • A participant notes that the original question has already received multiple responses on how to solve for x.

Areas of Agreement / Disagreement

Participants present multiple competing methods for solving the equation, and there is no consensus on a single approach. The discussion remains unresolved regarding the best method to express x in terms of y.

Contextual Notes

There are assumptions about the clarity of the original equation, and the discussion highlights the potential for misinterpretation without proper notation. The methods proposed depend on different algebraic manipulations and substitutions, which may lead to varying results.

gigi9
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Help please!

How can I change this y=2/3x^(3/2)-1/2x^(1/2) in term of y, as x=...?
 
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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).
 
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!)
 
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]
 
I have to solve the equation 2/3x^(3/2)-1/2x^(1/2) for "X", as x=...
 
Yes, that was what you said before and you got three replies telling you how to do that.
 
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.
 

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