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#1 2010-03-15 00:06:13
- ZHero
- Real Member
- Registered: 2008-06-08
- Posts: 1,889
Algebra Problem
has Real Roots. If a & b are Real then b can take any value EXCEPT .... ??
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#2 2010-03-15 11:30:02
- bobbym
- bumpkin
- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606
Re: Algebra Problem
Hi;
I am not sure what you are trying to restrict here. If you want 3 real roots then:
Δ > 0: the equation has 3 distinct real roots;
Δ < 0, the equation has 1 real root and 2 complex conjugate roots;
Δ = 0: at least 2 roots coincide, and they are all real.
Well for one thing b cannot equal 0 because then your poly will have 2 complex roots. But there are other b values that will yield complex roots also.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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#3 2010-03-15 15:32:10
- ZHero
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- Registered: 2008-06-08
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Re: Algebra Problem
How do you find
(Discriminant) for Cubic Equations??
Last edited by ZHero (2010-03-15 15:36:17)
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#4 2010-03-15 16:17:43
- bobbym
- bumpkin
- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606
Re: Algebra Problem
Hi;
The discriminant of the general cubic
Is:
They are computed using resultants and determinants. It is a difficult method and impossible by hand.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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#5 2010-03-15 18:33:03
- ZHero
- Real Member
- Registered: 2008-06-08
- Posts: 1,889
Re: Algebra Problem
Good!
That's Exactly what i was looking for!
Thanks 
If two or more thoughts intersect, there has to be a point!
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#6 2010-03-15 20:12:33
- bobbym
- bumpkin
- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606
Re: Algebra Problem
Hi ZHero;
I hope you meant the info on the discriminant because there is an efficient algorithm for determinants call the Fadeev Leverrier method. But it is designed for computers.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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#7 2010-03-16 16:30:11
- ZHero
- Real Member
- Registered: 2008-06-08
- Posts: 1,889
Re: Algebra Problem
mmmm... not exactly!
i meant, in our institute, b=2 was the Only solution known however i was pretty sure that other values existed too!
my approach was... let there be one Complex root, then Its Conjugate will also be a root!
let Complex Roots be A=m+n.i and B=m-n.i and C be the Real Root.
also SUM OF ROOTS=> A+B+C=2m+C=b/a
hence C=b/a-2m and f(C)=0 as C is a Root!
then.. getting a Quadratic in b and its Discriminant>=0 as b is Real and so on!
i was pretty sure that apart from b=2, Other Values of b also Existed and b=0 is one of them!!!
can u find ANY OTHER VALUE??
THX A LOT!
If two or more thoughts intersect, there has to be a point!
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#8 2010-03-16 18:02:36
- bobbym
- bumpkin
- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606
Re: Algebra Problem
Hi ZHero;
This is only a numerical estimate, but I believe that all values of b in the closed interval 0 to 2.7846 will have a discriminant < 0. There are other intervals that are dependent upon a.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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