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A cubic equation and its roots

  1. Apr 28, 2015 #1
    1. The problem statement, all variables and given/known data
    The question says that :
    Find the value of ##a## so that the equation $$x^3-6x^2+11x+a-6=0$$ has exactly three integer solitions.

    2. Relevant equations
    IF ##p##,##q##,##r## are the roots of this equation then:
    ##p+q+r=6##
    ##pq+pr+rq=11##
    ##pqr=6-a##

    3. The attempt at a solution
    I just don't know how to use these equations to find ##a##.
    So how to do this question?Are these equations of any use here?
     
  2. jcsd
  3. Apr 28, 2015 #2
    I would simply factor it.
     
  4. Apr 28, 2015 #3
    But how can you simply factor it when there is an unknown ##a## in the equation?
     
  5. Apr 28, 2015 #4

    haruspex

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    So factor the rest of it.
     
  6. Apr 28, 2015 #5

    Dick

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    So that's pretty much just guessing that ##a=0##, right? Or is there a better reason to do it? Suppose the constant were ##a-7##?
     
  7. Apr 28, 2015 #6

    haruspex

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    It just struck me as something worth doing to obtain insight. It doesn't presume a=0, in fact it reveals an infinity of solutions. A minor modification solves a-7.
     
  8. Apr 29, 2015 #7

    ehild

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    Find possible integer solutions which sum up to 6. Then select those for which pg+pr+qr=11.
    You can eliminate one of the roots. Try completing squares.
     
    Last edited: Apr 29, 2015
  9. Apr 29, 2015 #8
    Thank you....Did what you said.
    Leaving the zero values of ##p,q,r## , three possibilities were present:
    $$p=1,q=1,r=4$$, $$p=1,q=2,r=3$$,$$p=q=r=2$$
    Among which the second one satisfies the equation $$pq+qr+pr=11$$
    So ##pqr=6=6-a##
    Therefore ##a=0##.
    Is this how you meant it to be done?
     
  10. Apr 29, 2015 #9

    ehild

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    Yes, it was my first idea. There might be other solutions, including negative roots, but the problem text did not want you give or exclude more.
     
  11. Apr 29, 2015 #10

    haruspex

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    It says integer solutions, not natural numbers.
    Factorising the terms other than a makes it obvious what all the solutions are.
     
  12. Apr 29, 2015 #11

    Dick

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    I still don't see it. You aren't just saying that ##-a## is the product of three consecutive integers, are you? That suffices to give one integer root. Doesn't say there are three unless ##a=0##.
     
  13. Apr 29, 2015 #12

    haruspex

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    Ah yes, not thinking straight.
     
  14. Apr 29, 2015 #13

    SammyS

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    I have puzzled over how to use those equations for p, q, & r.

    Use the factor theorem, plus perhaps Descartes's Sign Rule.

    At most one negative root, that's if a > 6 . If -1 is a root, then a = 24.

    If 0 is a root, a = 6. Look at the resulting quadratic.

    If 1 is a root, a = __ , ...
     
  15. Apr 30, 2015 #14

    epenguin

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    You could use psychology. Why are they giving you the equation in the form

    ?

    The constant term is peculiar. It would be equivalent but normally more natural to ask what values of b give integral solutions of

    $$x^3-6x^2+11x+b=0$$ .

    It is because you might have to be a tad smarter to see how to do it when espressed that way. They are giving you a hint! I'll give you another. In the first equation the cubic without the a is quite easily factorised. Just knowing that it is makes it easier. Then if you think what a cubic looks like in a graph, you will see there is only a limited range of b within which it can have 3 real solutions and only a limited number of values for which it can have integer solutions.
     
    Last edited: May 1, 2015
  16. Apr 30, 2015 #15

    ehild

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    You meant factorising the part without a ?
     
  17. May 1, 2015 #16

    epenguin

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    Yes. Thank you, I have corrected that now.
     
  18. May 1, 2015 #17
    Okay then , now I understand that this equation can also be solved without trial and error method. So I did it as:
    ##x^3-6x^2+11x-6=-a##
    ## x^3-5x^2-x^2+5x+6x-6=-a##
    Which gives the equation:
    ##(x-1)(x-2)(x-3)+a=0##
    Which has integer roots only if ##a=0##
    Is that what you want me to understand?
     
  19. May 1, 2015 #18

    epenguin

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    Up to the last two lines. It has integer roots for a = 0 . The equation asks you find THE value of a that gives you that, so I suppose you have answered the question. You might worry a bit about whether this is the only possible value of a to give integer roots and how you would handle it if there were more than one value of a giving integer roots.

    Perhaps there are cases like that, say for a cubic. Perhaps you could construct such cases. Or show that there can't be any. Depending on your time and interest. Polya principle of 'How to solve it' - when you have solved it, it's not finished. But at least try see if you can see in this case if 0 is the only value.
     
  20. May 1, 2015 #19
    Okay I think I got what you want to say. We can have infinite values of ##a## to satisfy the equation. But since it's given that the equation has 3 roots therefore ##a## can have only one value that is ##0##.
    Is this what you want to say to me?
     
  21. May 1, 2015 #20

    epenguin

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    No :oldsmile: , I am saying it is not demonstrated and not all that obvious that there are no other values of a to give integral roots, and for someone wanting to develop a bit of mathematical muscle (of course you may consider you have better things to do with your time) it is of value to try and see further inside this.
     
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