Eigen Values for k^2-8k+-12=0: (k-2) and (k+3)

In summary, the conversation discussed the process of finding eigenvalues for a quadratic equation and the correct interpretation of the equation k^2-8k+-12=0. The participants also clarified the difference between using the quadratic formula and factoring the equation. It was emphasized that factoring is only possible if the equation is in the form k^2+bk+c.
  • #1
what are the eigen values for


i got (k-2) (k+3) as eigen values
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  • #2
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.
  • #3
its just the quadratic formula in my book.i think k is a integra to be worked out.
  • #4
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?
  • #5
the factorization.
  • #6
i think it is k^2-(a+b)k+ad-bc=0
  • #7
morbello said:
what are the eigen values for


i got (k-2) (k+3) as eigen values

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.
  • #8
so may be it is (k-2)(k-3) and muliti ply by 4
  • #9
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.
  • #10
No, then your leading order term would be [tex]4k^2[/tex]
  • #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
  • #12
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?
  • #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
  • #14
morbello said:

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.
  • #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
  • #16
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.
  • #17
ok thank you ill try it .im an adult learner so I am trying to understand what was done in alevel maths .thank you again.
  • #18
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].
  • #19
thank you yet again sorry to have had you trying to work out what i was trying to portray

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