Solving for x in y=2x-3/(2x-4)^2: Tips and Hints for Solving Equations

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Homework Help Overview

The discussion revolves around solving the equation y=2x-3/(2x-4)^2 for the variable x. Participants are exploring the complexities of isolating x within this equation, which involves polynomial expressions and potentially cubic equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants suggest multiplying both sides by (2x-4)^2 to eliminate the fraction, leading to a cubic equation. Some express confusion about how to isolate x after this transformation.

Discussion Status

There is ongoing exploration of the equation's structure, with some participants noting the emergence of a cubic form and discussing the implications of having y as a coefficient. While various attempts to simplify the equation are shared, there is no clear consensus on a straightforward method to isolate x.

Contextual Notes

Some participants mention the complexity of cubic equations and the potential involvement of complex numbers, indicating that the problem may not have a simple resolution. There are also references to earlier posts that highlight the distinction between quadratic and cubic forms in similar equations.

e^(i Pi)+1=0
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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-\frac{3}{(2x-4)^2}
 
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Multiply both sides of the equation by (2x- 4)^2 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
 
But then I'm left with y(2x-4)^2 on the left hand side.
 
The cubic factors to 8(x-1.5)(x-\frac{5+\sqrt{21}}{4})(x-\frac{5-\sqrt{21}}{4})
 
e^(i Pi)+1=0 said:
But then I'm left with y(2x-4)^2 on the left hand side.

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.
 
e^(i Pi)+1=0 said:
But then I'm left with y(2x-4)^2 on the left hand side.
Yes, you do. So what? y will be part of the coefficients of the polynomial so the solution will involve y.
 
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.
 
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...
 
e^(i Pi)+1=0 said:
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.


You cannot isolate x by such methods. It involves complex numbers, which is described in HallsOfIvy's link. :smile:
 
  • #10
e^(i Pi)+1=0 said:
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.


Somehow you are not "getting it", so let's look at a simpler case:
y = 2x - \frac{3}{2x-4}.
If you multiply through by the denominator you get (2x-4)y = 4x^2-8x-3, or
4x^2 -(8+2y)x + (4y-3) = 0, 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:
x = 1 + \frac{y}{4} \pm \frac{\sqrt{y^2 - 8y + 28}}{4}.

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
 
  • #11
Thank you! I didn't recognize that there was a quadratic hidden in there.
 
  • #12
e^(i Pi)+1=0 said:
Thank you! I didn't recognize that there was a quadratic hidden in there.
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:
  • #13
Yeah, I get it now, the Xs factor out to make a cubic. I was just full of fail this thread.
 
  • #14
e^(i Pi)+1=0 said:
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-\frac{3}{(2x-4)^2}

<br /> \frac{3}{(2 x - 4)^2} = 2x - y<br />
<br /> 3 = (2 x - 4)^2 (2 x -y)<br />
<br /> 3 = 4 (2 x - y) (x^2 - 4 x + 4)<br />
<br /> 3 = 8 x^3 - 4 (y + 8) x^2 + 16 ( y + 2) \, x - 16 y<br />

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

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:
  • #15
Dickfore said:
<br /> \frac{3}{(2 x - 4)^2} = 2x - y<br />
<br /> 3 = (2 x - y)^2 (2 x -y)<br />
Umm... no.
 
  • #16
Dickfore said:
<br /> 3 = (2 x - 4)^2 (2 x -y)<br />
<br /> 3 = 4 (2 x - y) (x^2 - 4 x + 4)<br />

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

EDIT: My mistake, this is correct. :redface:
 
Last edited:
  • #17
oay said:
Keep trying...! :smile:

I believe that step is correct.
 
  • #18
Dickfore said:
I believe that step is correct.
Bugger me. My sincerest apologies! :redface: You are of course correct.

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

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