1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Equation homework problem

  1. Jun 28, 2005 #1
    Solve for x:
    I got the final equation but I am unable to get the it form there.
  2. jcsd
  3. Jun 28, 2005 #2


    User Avatar
    Gold Member

    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
  4. Jun 28, 2005 #3
    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
  5. Jun 28, 2005 #4
    [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
  6. Jun 28, 2005 #5
    You made a typo, [tex]x^4+2x^3-x^2-6x-3 = (x^2-x-1)(x^2+3x+3)[/tex]
  7. Jun 28, 2005 #6
    Yes, of course!

    it's [tex]x^2-x-1[/tex]
  8. Jun 29, 2005 #7
    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
  9. Jun 29, 2005 #8
    Is there a actual way to solve these type of sums! PLS I NEED HELP I AM DUMB.
  10. Jun 29, 2005 #9


    User Avatar
    Science Advisor
    Homework Helper

    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:


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


    Can you make that substitution?

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


    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":


    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:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook