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

[morbello]

what are the eigen values for


k^2-8k+-12=0

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

 

[matt grime]

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.

 

[morbello]

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

 

[matt grime]

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?

 

[morbello]

the factorization.

 

[morbello]

i think it is k^2-(a+b)k+ad-bc=0

 

[jbunniii]

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.

 

[morbello]

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

 

[Cyosis]

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.

 

[Born2bwire]

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

 

[morbello]

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

 

[Cyosis]

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?

 

[morbello]

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

 

[Cyosis]

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.

 

[morbello]

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

 

[Cyosis]

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.

 

[morbello]

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.

 

[Cyosis]

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 $c=\alpha \beta$.

 

[morbello]

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