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Solving in terms of x

  1. Jun 16, 2012 #1
    How would I solve this for x? I've been trying for a while now, and
    i can't seem to get anywhere. I just need some kind of hint.

    y=2x-[itex]\frac{3}{(2x-4)^2}[/itex]
     
  2. jcsd
  3. Jun 16, 2012 #2

    HallsofIvy

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    Multiply both sides of the equation by [itex](2x- 4)^2[/itex] and you will have a cubic equation for x. There is a formula for the solution of general cubic polynomial equations but it is very complicated.

    http://en.wikipedia.org/wiki/Cubic_function
     
  4. Jun 16, 2012 #3
    But then I'm left with y(2x-4)^2 on the left hand side.
     
  5. Jun 16, 2012 #4
    The cubic factors to 8(x-1.5)(x-[itex]\frac{5+\sqrt{21}}{4}[/itex])(x-[itex]\frac{5-\sqrt{21}}{4}[/itex])
     
  6. Jun 16, 2012 #5
    I just gave it a shot.
    Expand y(2x-4)^2 and take everything on one side, you will get a cubic equation, solve it taking x as variable. This will take a little time, there must be a simpler way too.
     
  7. Jun 16, 2012 #6

    HallsofIvy

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    Yes, you do. So what? y will be part of the coefficients of the polynomial so the solution will involve y.
     
  8. Jun 16, 2012 #7
    We have 0=8x3-32x2+32x-3-4x2y+16xy-16y

    Here are some forms I've managed to simplify it to

    0=(x-2)2(8x-4y)-3

    8x(x-2)2=4y(x-2)2+3

    I just cannot seem to isolate x.
     
  9. Jun 16, 2012 #8
    If you really want to solve for x, you'll have to use the method shown in the wiki link Halls gave earlier. Wolframalpha gave a complicated-looking answer...
     
  10. Jun 17, 2012 #9
    You cannot isolate x by such methods. It involves complex numbers, which is described in HallsOfIvy's link. :smile:
     
  11. Jun 17, 2012 #10

    Ray Vickson

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    Somehow you are not "getting it", so let's look at a simpler case:
    [tex] y = 2x - \frac{3}{2x-4}.[/tex]
    If you multiply through by the denominator you get [itex](2x-4)y = 4x^2-8x-3,[/itex] or
    [itex] 4x^2 -(8+2y)x + (4y-3) = 0,[/itex] so you get a quadratic equation for x with "y" in some of the coefficients. That means that the solution will be a reasonably complicated function of y:
    [tex] x = 1 + \frac{y}{4} \pm \frac{\sqrt{y^2 - 8y + 28}}{4}.[/tex]

    The same type of thing happens in your original problem, but now you end up with a cubic equation for x (with y in some coefficients), so the final solutions for x are much, much more complicated.

    RGV
     
  12. Jun 17, 2012 #11
    Thank you! I didn't recognize that there was a quadratic hidden in there.
     
  13. Jun 17, 2012 #12
    There isn't!

    RV's simpler example involved a quadratic, but your original problem involves a cubic, as was pointed out to you by Ivy in post #2.

    (BTW, your username is my favourite equation ever.)
     
    Last edited: Jun 17, 2012
  14. Jun 17, 2012 #13
    Yeah, I get it now, the Xs factor out to make a cubic. I was just full of fail this thread.
     
  15. Jun 17, 2012 #14
    [tex]
    \frac{3}{(2 x - 4)^2} = 2x - y
    [/tex]
    [tex]
    3 = (2 x - 4)^2 (2 x -y)
    [/tex]
    [tex]
    3 = 4 (2 x - y) (x^2 - 4 x + 4)
    [/tex]
    [tex]
    3 = 8 x^3 - 4 (y + 8) x^2 + 16 ( y + 2) \, x - 16 y
    [/tex]

    You get the following cubic equation:
    [tex]
    8 x^3 - 4 (y + 8) x^2 + 16 ( y + 2) \, x - (16 y + 3) = 0
    [/tex]

    EDIT:
    I made a typo and gave a wrong equation. I believe it is corrected now.

    Do you know how to solve cubic equations?
     
    Last edited: Jun 18, 2012
  16. Jun 17, 2012 #15
    Umm... no.
     
  17. Jun 18, 2012 #16
    Keep trying...! :smile:

    EDIT: My mistake, this is correct. :redface:
     
    Last edited: Jun 18, 2012
  18. Jun 18, 2012 #17
    I believe that step is correct.
     
  19. Jun 18, 2012 #18
    Bugger me. My sincerest apologies! :redface: You are of course correct.

    (I went blind for a second and misread you.)
     
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