complex algebra

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• June 29th 2010, 08:55 AM
Dinkydoe
Quote:

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.
• June 29th 2010, 11:18 PM
hooke
Quote:

Originally Posted by Dinkydoe
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 !
• June 30th 2010, 02:13 AM
hooke
Quote:

Originally Posted by hooke
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 ?
• June 30th 2010, 03:55 AM
Dinkydoe
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 or $m>4$
• June 30th 2010, 11:27 PM
ysye
it is very easy,but your thinking is worry. wow gold
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