# Equation homework problem

1. Jun 28, 2005

### stupidkid

PHP:
Solve for x:
x^2+x^2/(x+1)^2=3
I got the final equation but I am unable to get the it form there.
x^4-x^2+2x^3-6x-3=0

2. Jun 28, 2005

### cronxeh

I'm afraid there is no simple way to find x - you'll have to use numerical methods. You can find first 2 solutions by graphing this, and the last 2 are complex. I've enclosed the graph that should help you find where your roots are located for real numbers

Last edited: Oct 8, 2005
3. Jun 28, 2005

### Maxos

It is incorrect, there is a formula to solve algebraic eqations until the 4 grade.

I have links to italian pages, but, if you try you'll find something in english, too.

The two real solutions are (1+-sqrt(5))/2

Last edited: Jun 28, 2005
4. Jun 28, 2005

### Popey

$$x^4+2x^3-x^2-6x-3 = (x^2-x+1)(x^2+3x+3)$$

The first has the solutions above
The second has no real solutions

5. Jun 28, 2005

### Maxos

You made a typo, $$x^4+2x^3-x^2-6x-3 = (x^2-x-1)(x^2+3x+3)$$

6. Jun 28, 2005

### Popey

Yes, of course!

it's $$x^2-x-1$$

7. Jun 29, 2005

### geniusprahar_21

there is another method to do it though using quadratic equations entirely.
use a2 + b2 = (a - b)2 + 2ab. then put x2/(x+1) = y

8. Jun 29, 2005

### stupidkid

Is there a actual way to solve these type of sums! PLS I NEED HELP I AM DUMB.

9. Jun 29, 2005

### saltydog

You know, there is a method for solving these "quartic" equations. It's messy and confussing though. I'll describe it in general terms; you can investigate it further if you wish. Consider the general case:

$$x^4+bx^3+cx^2+dx+e=0$$

We can make a substitution $x=y-\frac{1}{4}b$ to convert it to a "reduced quartic":

$$y^4+qy^2+ry+s=0$$

Can you make that substitution?

Now, assume the reduced quartic can be factored into two quadratics:

$$y^4+qy^2+ry+s=(y^2+2ky+l)(y^2-2ky+m)$$

where k,l, and m are to be determined. You can equate coefficients of like powers and end up with two equations for l and m in terms of k. Upon making some additional substitutions you'll end up with a sixth-degree polynomial called the "resolvent cubic":

$$64k^6+32qk^4+4(q^2-4s)k^2-r^2=0$$

This resolvent cubic is actually a cubic equation in $k^2$. Now, you can use the method, equally messy, to solve for a general cubic equation for [itex]k^2[/tex] and then use one value of k to determine l and m, then the quadratics above.

You're thinking now, "forget that dude, I'll just plug it into Mathematica and call it done".