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I want answers

  1. Jul 15, 2011 #1
    [URL]http://www13.0zz0.com/2011/07/15/22/754178270.jpg[/URL]

    I Need help solving this fourth degree equation
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Jul 15, 2011 #2

    Pengwuino

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    "I want answers" is not how you're going to get results. Read the rules of the forums.
     
  4. Jul 15, 2011 #3

    micromass

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    2016 Award

  5. Jul 15, 2011 #4
    Me, too.
     
  6. Jul 15, 2011 #5
    lol comic sans
     
  7. Jul 16, 2011 #6
    thank you

    Finally :smile: I found a solution

    301263258.jpg
     
  8. Jul 16, 2011 #7

    I like Serena

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    Hmm.
    I understand the solution x=-1.
    But how did you get the other solutions from [itex]9x^3-4x+1=0[/itex]? :confused:
     
  9. Jul 16, 2011 #8

    HallsofIvy

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    The simplest way to approach this problem is to use the "rational root theorem".

    If r is a rational number satisfying [itex]a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0[/itex] where the coefficients are all integers, then r= p/q where p is an integer evenly dividing [itex]a_0[/itex] and q is an integer evenly dividing [itex]a_n[/itex].

    Of course, it is not necessary than such an equation have any rational roots but it is worth trying. Here, the leading coefficient, [itex]a_n[/itex], is 9, which has factors [itex]\pm 1, \pm 3, \pm 9[/itex] and the constant term, [itex]a_0[/itex] is 1, which has factors [itex]\pm 1[/itex] so the only possible rational roots are [itex]\pm 1, \pm 1/3, \pm 1/9[/itex].

    Putting those into the equation, we see that if
    [tex]9(1)^4+ 9(1)^3- 4(1)^2- 3(1)+ 1= 19- 7= 12\ne 0[/tex]
    [tex]9(-1)^4+ 9(-1)^3- 4(-1)^2- 3(-1)+ 1= -9+ 9- 4+ 3+ 1= -1+1= 0[/tex]
    [tex]9(1/3)^4+ 9(1/3)^3- 4(1/3)^2- 3(1/3)+ 1= 1/9+ 1/3- 4/9- 1+ 1= 1/3- 1/3- 1+ 1= 0[/tex]
    We can stop here. Seeing that x= -1 and x= 1/3 are roots, we can divide by x+ 1 and x- 1/3 (not "0.3333") to get a quadratic equation that we can solve using the quadratic formula.

    (There is a "quartic formula", http://www.sosmath.com/algebra/factor/fac12/fac12.html, but it is extermely complicated.)
     
  10. Jul 16, 2011 #9

    I like Serena

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    Nice! I didn't know this one yet! :smile:
     
  11. Jul 16, 2011 #10
    this is another solution

    648322002.jpg

    Is this true ?
     
  12. Jul 16, 2011 #11

    Redbelly98

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    Moderator's note: thread moved from "General Math".

    Please help us moderate the forum by reporting misplaced homework posts.

    Thank you.
     
  13. Jul 16, 2011 #12

    Redbelly98

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    Why don't you tell us? Are those answers the same as the ones you already know:
     
  14. Jul 16, 2011 #13
    yeah,I just want clarification
     
  15. Jul 16, 2011 #14

    I like Serena

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    What do you think?
    Can you think of a way to derive this solution?

    Btw, are you familiar with solving a quadratic equation?
    That is, an equation of the form ax2 + bx + c = 0.
     
  16. Jul 16, 2011 #15
    Yes, do you mean you want the law of quadratic equation
     
  17. Jul 16, 2011 #16

    I like Serena

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    What I would like is for you to give us some insight into what you are thinking.
    That would make it easier for us to help you.

    As it is we have no clue how you got this problem or why you want to solve it.
    I'm assuming you're supposed to learn how to solve such equations.
    At least that's what we're trying to help you with.

    How did you get this problem?
    And what is your purpose with it?
     
  18. Jul 16, 2011 #17
    thank you ( i like serena ) for help me .
     
  19. Jul 17, 2011 #18
    Dude! This is an awesome theorem, thanks for that.
     
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