Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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