• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Equation homework problem

  • Thread starter stupidkid
  • Start date
18
0
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
 

cronxeh

Gold Member
949
10
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:
92
0
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:
22
0
[tex]x^4+2x^3-x^2-6x-3 = (x^2-x+1)(x^2+3x+3)[/tex]

The first has the solutions above
The second has no real solutions
 
92
0
You made a typo, [tex]x^4+2x^3-x^2-6x-3 = (x^2-x-1)(x^2+3x+3)[/tex]
 
22
0
Yes, of course!

it's [tex]x^2-x-1[/tex]
 
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
 
18
0
Is there a actual way to solve these type of sums! PLS I NEED HELP I AM DUMB.
 

saltydog

Science Advisor
Homework Helper
1,582
2
stupidkid said:
Is there a actual way to solve these type of sums! PLS I NEED HELP I AM DUMB.
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:

[tex]x^4+bx^3+cx^2+dx+e=0[/tex]

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

[tex]y^4+qy^2+ry+s=0[/tex]

Can you make that substitution?

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

[tex]y^4+qy^2+ry+s=(y^2+2ky+l)(y^2-2ky+m)[/tex]

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":

[tex]64k^6+32qk^4+4(q^2-4s)k^2-r^2=0[/tex]

This resolvent cubic is actually a cubic equation in [itex]k^2[/itex]. 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". :smile:
 

Related Threads for: Equation homework problem

  • Posted
Replies
2
Views
987
  • Posted
Replies
3
Views
14K
  • Posted
Replies
5
Views
2K
Replies
6
Views
16K
  • Posted
Replies
11
Views
2K
  • Posted
Replies
7
Views
1K
  • Posted
Replies
3
Views
2K
  • Posted
Replies
2
Views
1K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top