Homework Help: Solving in terms of x

1. Jun 16, 2012

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

2. Jun 16, 2012

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

3. Jun 16, 2012

e^(i Pi)+1=0

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

4. Jun 16, 2012

e^(i Pi)+1=0

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

5. Jun 16, 2012

Saitama

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.

6. Jun 16, 2012

HallsofIvy

Yes, you do. So what? y will be part of the coefficients of the polynomial so the solution will involve y.

7. Jun 16, 2012

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.

8. Jun 16, 2012

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...

9. Jun 17, 2012

Infinitum

You cannot isolate x by such methods. It involves complex numbers, which is described in HallsOfIvy's link.

10. Jun 17, 2012

Ray Vickson

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. Jun 17, 2012

e^(i Pi)+1=0

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

12. Jun 17, 2012

skiller

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.

Last edited: Jun 17, 2012
13. Jun 17, 2012

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.

14. Jun 17, 2012

Dickfore

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

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?

Last edited: Jun 18, 2012
15. Jun 17, 2012

skiller

Umm... no.

16. Jun 18, 2012

skiller

Keep trying...!

EDIT: My mistake, this is correct.

Last edited: Jun 18, 2012
17. Jun 18, 2012

Dickfore

I believe that step is correct.

18. Jun 18, 2012

skiller

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

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