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Math Help - complex algebra

  1. #16
    Senior Member Dinkydoe's Avatar
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    Can \alpha=\beta ?I thought its stated clearly that

    \alpha<-1<\beta
    No, they're not equal at all. The only thing we change is 'what' we call \alpha,\beta

    We call the roots, \alpha,\beta with \alpha < -1 < \beta

    We simply call \alpha, the smallest of the 2 roots, and \beta the other. Thus if the inequality '<' flips into '>' due to a negative sign, then we only change what we call \alpha,\beta . Perhaps, I was unclear.

    I hope you understand what I mean, now.
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  2. #17
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    Quote Originally Posted by Dinkydoe View Post
    No, they're not equal at all. The only thing we change is 'what' we call \alpha,\beta

    We call the roots, \alpha,\beta with \alpha < -1 < \beta

    We simply call \alpha, the smallest of the 2 roots, and \beta the other. Thus if the inequality '<' flips into '>' due to a negative sign, then we only change what we call \alpha,\beta . Perhaps, I was unclear.

    I hope you understand what I mean, now.
    yeah, they are unknowns anyways. No worries Dinky, you've been very helpful and precise in explaining. thanks !
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  3. #18
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    Quote Originally Posted by hooke View Post
    yeah, they are unknowns anyways. No worries Dinky, you've been very helpful and precise in explaining. thanks !
    I would like to do a final checking, is the answer -1<m<4 ?
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  4. #19
    Senior Member Dinkydoe's Avatar
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    I found something slightly different.

    I guess that you also found that

    \Delta(f)>(-m^2+6m+10)^2

    could be simplified to 12m^3-60m^2+24m+96> 0

    This polynomial can be factored:

    g(m)=12(m+1)(m-4)(m-2). Thus we find zero's m=-1,m=2,m=4

    Then we can argue that, (or see by plotting) that g(m)> 0 as -1<m<2 or m>4
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  5. #20
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    it is very easy,but your thinking is worry. wow gold
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