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Proove there is an x s.t. x^3+x=6

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that there is an x∈R such that x^3+x=6.

    3. The attempt at a solution

    I'm not exactly sure where to get started with this proof. I think I would need to define a set S={x∈R: x>0 and x^3+x≤6}. Assume S is bounded, and then find lub(S)...?
     
  2. jcsd
  3. Sep 13, 2009 #2

    jgens

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    Gold Member

    I think with some work, your approach will definately yield a correct proof; however, are you familiar with the Intermediate Value Theorem?
     
  4. Sep 13, 2009 #3
    There are 3 solutions, if you prove one is complex then one must be real?
     
  5. Sep 13, 2009 #4
    How exactly would I use the Intermediate Value Theorem?
     
  6. Sep 13, 2009 #5
    Would I need to first prove that it is continuous then?
     
  7. Sep 13, 2009 #6
    All polynomial functions are continuous over all of R.

    The Intermediate Value Theorem is used frequently to establish that a solution to an equation exists (but oftern is no help in finding that solution, except by numeric approximation methods).

    As an example: Show there is a solution to the equation cos(x) = x.

    This is equivalent to showing that cos(x) - x = 0 is solvable.

    Since f(x) = cos(x) - x is a continuous function and f(0) = 1 > 0 and f(pi/2) = -pi/2 < 0 there must exist some value c between 0 and pi/2 where f(c) = 0 (by the Intermediate Value Theorem), hence there is a solution to cos(x) = x.

    Your problem can be solved similarly.

    --Elucidus
     
  8. Sep 14, 2009 #7
    And why don't you simply solve x3+x=6 and show that there are real values for x.

    [tex]x^3+x-6=0[/tex]

    Start by checking if [itex]\pm 1, \pm 2, \pm 3, \pm 6[/itex] are solutions of the cubic equation.
     
  9. Sep 14, 2009 #8

    HallsofIvy

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    Well, unless there is an integer root, that won't "solve" the equation! But that is a good way to start. In fact, it is exactly what jgens suggested. If x= 0 [itex]x^3+ x= 0+ 0= 0< 6[/itex]. If x= 2, [itex]x^3+ x= 8+ 2= 10> 6[/itex]. What does that tell you?
     
  10. Sep 14, 2009 #9
    It tells me that the solution is between 0 and 2, i.e [itex]x\in (0,2)[/itex].

    Infact if you choose x=1, [itex]1^3+1=1+1=2<6[/itex], so [itex]x \in (1,2)[/itex]

    If you go by checking for x=1.5 , [itex]1.5^3+1.5=4.875<6[/itex] you will come up with the solution. Now [itex]x \in (1.5 , 2)[/itex] and so on...

    Edit: In addition, I came up with better solution.

    Check the sign of the discriminant of the cubic equation:

    [tex]D=\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}[/tex]

    The general cubic equation is:

    [tex]x^3+px+q[/tex]

    Just extract p and q from your equation.

    1)If D>0 then there are 1 real and two conjugate complex roots.

    2)If D=0 there is one prime and two equal real roots.

    3)If D<0 there are 3 real and different roots.
     
    Last edited: Sep 14, 2009
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