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

• morbello
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.

#### morbello

what are the eigen values for

k^2-8k+-12=0

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

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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.

its just the quadratic formula in my book.i think k is a integra to be worked out.

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?

the factorization.

morbello said:
what are the eigen values for

k^2-8k+-12=0

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

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

$$k^2 - 8k - 12 = 0$$

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.

so may be it is (k-2)(k-3) and muliti ply by 4

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.

No, then your leading order term would be $$4k^2$$

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

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?

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

morbello said:
k^2-8k+-12=0

Most people would read this as either $k^2-8k-12=0$ or $k^2-8k \pm 12=0$. The first one, $k^2-8k-12=0$, cannot be factored by using integers. However, $k^2-8k+12=0$ 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.

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

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 $k^2-8k-12=0$. This equation cannot be factorized and you will need to use the ABC formula.

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.

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

Secondly factorizing a quadratic equation $k^2+bk+c$ can only be done if you can write it in this form, $k^2+(\alpha+\beta)k+\alpha \beta$, with $b=(\alpha+\beta)$ and $c=\alpha \beta$.

thank you yet again sorry to have had you trying to work out what i was trying to portray