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Eigen vaules.

  1. May 22, 2009 #1
    what are the eigen values for


    k^2-8k+-12=0

    i got (k-2) (k+3) as eigen values
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: May 22, 2009
  2. jcsd
  3. May 22, 2009 #2

    matt grime

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    What is K, and what is k? Endomorphisms of vector spaces or (square) matrices have eigenvalues. You're implying that k and K are scalars.
     
  4. May 22, 2009 #3
    its just the quadratic formula in my book.i think k is a integra to be worked out.
     
  5. May 22, 2009 #4

    matt grime

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    Presumably you mean that you have a 2x2 matrix M and working out det(M-kI) yields the quadratic equation


    k^2-8k +12 (or possibly k^2-8k-12)

    and you want to find the roots of this quadratic equation, or its factorization.

    Is that the correct interpretation of what you're saying?
     
  6. May 22, 2009 #5
    the factorization.
     
  7. May 22, 2009 #6
    i think it is k^2-(a+b)k+ad-bc=0
     
  8. May 22, 2009 #7

    jbunniii

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    Are you saying that the characteristic equation for some linear map is

    [tex]k^2 - 8k - 12 = 0[/tex]

    and you factored the polynomial as (k-2)(k+3)?

    If so, that's incorrect, as you can see by simply multiplying (k-2)(k+3) and seeing that you don't get back your original polynomial.
     
  9. May 22, 2009 #8
    so may be it is (k-2)(k-3) and muliti ply by 4
     
  10. May 22, 2009 #9

    Cyosis

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    Don't randomly guess and work out the brackets instead. You will see that multiplying by four will do no good. Are you familiar with the ABC formula if so use it.
     
  11. May 22, 2009 #10

    Born2bwire

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    No, then your leading order term would be [tex]4k^2[/tex]
     
  12. May 22, 2009 #11
    i understand what you have said as you have to mutiply all the equation but i keep getting stuck on (k+2) (k-12) not understanding the part once you have the 8 to 12 by adding 2,then you take away 12 to equal 0
     
  13. May 22, 2009 #12

    Cyosis

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    Are you aware that not all quadratic equations can be factored in such a neat form? In fact most cannot, this one cannot for -12. It can however for +12, which one are you trying to calculate?
     
  14. May 22, 2009 #13
    im still having trouble with factorization i guess more than know how to do it have you a way that explains it easyer.i know you have to get both number to add up to the same or subtract to make the anwser 0
     
  15. May 22, 2009 #14

    Cyosis

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    Most people would read this as either [itex]k^2-8k-12=0[/itex] or [itex]k^2-8k \pm 12=0[/itex]. The first one, [itex]k^2-8k-12=0[/itex], cannot be factored by using integers. However, [itex]k^2-8k+12=0[/itex] can. Which equation you are trying to solve still isn't clear after 13 posts. And I am not the first one to ask you so please read the posts carefully.
     
  16. May 22, 2009 #15
    as it has said k^2-8k+-12=0 were the -12 is added to the equation would this not work like the k^2-8k+12=0
     
  17. May 22, 2009 #16

    Cyosis

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    Yes you have listed that equation which has a certain inherent ambiguity. I explained this in my previous post. From your last post the equation you're trying to solve is [itex]k^2-8k-12=0[/itex]. This equation cannot be factorized and you will need to use the ABC formula.
     
  18. May 22, 2009 #17
    ok thank you ill try it .im an adult learner so im trying to understand what was done in alevel maths .thank you again.
     
  19. May 22, 2009 #18

    Cyosis

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    If you're new to this I really suggest that you use the ABC formula on both [itex]k^2-8k-12=0[/itex] and [itex]k^2-8k+12=0[/itex]. This way you will see how different the answers are.

    Secondly factorizing a quadratic equation [itex]k^2+bk+c[/itex] can only be done if you can write it in this form, [itex]k^2+(\alpha+\beta)k+\alpha \beta[/itex], with [itex]b=(\alpha+\beta)[/itex] and [itex]c=\alpha \beta[/itex].
     
  20. May 22, 2009 #19
    thank you yet again sorry to have had you trying to work out what i was trying to portray
     
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