# Help !

1. Nov 30, 2003

### gigi9

Help plz!

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

2. Nov 30, 2003

### 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).

3. Dec 1, 2003

### HallsofIvy

Staff Emeritus
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))).

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!)

4. Dec 1, 2003

### 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,

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

5. Dec 1, 2003

### gigi9

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

6. Dec 1, 2003

### HallsofIvy

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

7. Dec 1, 2003

### StephenPrivitera

I always assume calculator syntax if it's unclear.