- #1

- 73

- 0

what are the eigen values for

k^2-8k+-12=0

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

k^2-8k+-12=0

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

## Homework Statement

## Homework Equations

## The Attempt at a Solution

Last edited:

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter morbello
- Start date

- #1

- 73

- 0

k^2-8k+-12=0

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

Last edited:

- #2

matt grime

Science Advisor

Homework Helper

- 9,420

- 4

- #3

- 73

- 0

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

- #4

matt grime

Science Advisor

Homework Helper

- 9,420

- 4

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

- 73

- 0

the factorization.

- #6

- 73

- 0

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

- #7

- 3,474

- 257

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

[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

- 73

- 0

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

- #9

Cyosis

Homework Helper

- 1,495

- 0

- #10

Born2bwire

Science Advisor

Gold Member

- 1,779

- 22

No, then your leading order term would be [tex]4k^2[/tex]

- #11

- 73

- 0

- #12

Cyosis

Homework Helper

- 1,495

- 0

- #13

- 73

- 0

- #14

Cyosis

Homework Helper

- 1,495

- 0

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

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],

- #15

- 73

- 0

- #16

Cyosis

Homework Helper

- 1,495

- 0

- #17

- 73

- 0

- #18

Cyosis

Homework Helper

- 1,495

- 0

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

- 73

- 0

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

Share: