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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

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## The Attempt at a Solution

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- Thread starter morbello
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- #1

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k^2-8k+-12=0

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

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- #2

matt grime

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

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matt grime

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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?

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the factorization.

- #6

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

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Are you saying that the characteristic equation for some linear map iswhat are the eigen values for

k^2-8k+-12=0

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

[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.

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

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Cyosis

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Born2bwire

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

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Cyosis

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

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Cyosis

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- #18

Cyosis

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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].

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

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