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

[e^(i Pi)+1=0]

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-$\frac{3}{(2x-4)^2}$

 

[HallsofIvy]

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

 

[e^(i Pi)+1=0]

But then I'm left with y(2x-4)^2 on the left hand side.

 

[e^(i Pi)+1=0]

The cubic factors to 8(x-1.5)(x-$\frac{5+\sqrt{21}}{4}$)(x-$\frac{5-\sqrt{21}}{4}$)

 

[Saitama]

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.

 

[HallsofIvy]

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.

 

[e^(i Pi)+1=0]

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.

 

[Bohrok]

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...

 

[Infinitum]

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.

 

[Ray Vickson]

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

 

[e^(i Pi)+1=0]

Thank you! I didn't recognize that there was a quadratic hidden in there.

 

[skiller]

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.)

 

[e^(i Pi)+1=0]

Yeah, I get it now, the Xs factor out to make a cubic. I was just full of fail this thread.

 

[Dickfore]

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-$\frac{3}{(2x-4)^2}$

$$\frac{3}{(2 x - 4)^2} = 2x - y$$
$$3 = (2 x - 4)^2 (2 x -y)$$
$$3 = 4 (2 x - y) (x^2 - 4 x + 4)$$
$$3 = 8 x^3 - 4 (y + 8) x^2 + 16 ( y + 2) \, x - 16 y$$

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

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

Do you know how to solve cubic equations?

 

[skiller]

$$\frac{3}{(2 x - 4)^2} = 2x - y$$
$$3 = (2 x - y)^2 (2 x -y)$$

Umm... no.

 

[skiller]

$$3 = (2 x - 4)^2 (2 x -y)$$
$$3 = 4 (2 x - y) (x^2 - 4 x + 4)$$

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

Keep trying...!

EDIT: My mistake, this is correct.

 

[Dickfore]

Keep trying...!

I believe that step is correct.

 

[skiller]

I believe that step is correct.

Bugger me. My sincerest apologies! You are of course correct.

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