How many negative roots has the equation

$\displaystyle x^6 - bx^5 - 2ax^3 - cx + a^2=0, b\geq0, c\geq0?$

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Malay

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- Jul 16th 2006, 08:16 PMmalaygoelRoots of equation
How many negative roots has the equation

$\displaystyle x^6 - bx^5 - 2ax^3 - cx + a^2=0, b\geq0, c\geq0?$

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Malay - Jul 17th 2006, 10:00 AMgalactus
Since $\displaystyle f(-x)=x^{6}+bx^{5}+2ax^{3}+cx+a^{2}$ has no variation in signs, there are no negative roots.

It's been awhile since I thought about Descartes rule. Isn't that the way it goes?. - Jul 17th 2006, 10:50 AMQuickQuote:

Originally Posted by**galactus**

$\displaystyle f(-x)=-x^{6}+bx^{5}+2ax^{3}+cx+a^{2}$

which does have a variation. - Jul 17th 2006, 02:17 PMCaptainBlackQuote:

Originally Posted by**galactus**

powers up to the maximum must be non-zero.

RonL - Jul 17th 2006, 02:51 PMgalactus
That would be true, Cap'n. I forgot about that. In Descartes rule, it is assumed that terms with 0 coefficients are deleted and the constant term is not 0. Also, descartes rule says the number of negative roots is equal to the change in signs or less than that by an even integer.

Let's say b and c are 0, then we'd have $\displaystyle (-x)^{6}+2a(-x)^{3}+a^{2}=x^{6}-2ax^{3}+a^{2}$

2 change of signs. Has 2 negative roots or 0 negative roots.

If b and c were not 0 , then from before. No change of signs. No negative zeros.

BTW, Quick, $\displaystyle (-x)^{6}=x^{6}$. May I ask where you got the negative from?. Maybe I am missing something. - Jul 17th 2006, 04:05 PMQuickQuote:

Originally Posted by**galactus**

- Jul 17th 2006, 07:38 PMmalaygoel
I am not familiar with Descartes rule. Please guide me.

What is the answer?What is the use of f(-x)?

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Malay - Jul 17th 2006, 10:54 PMCaptainBlackQuote:

Originally Posted by**malaygoel**

RonL - Jul 18th 2006, 02:26 AMmalaygoelQuote:

Originally Posted by**CaptainBlack**

What is the logic behind Descartes rule?

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Malay - Jul 18th 2006, 02:27 AMmalaygoelQuote:

Originally Posted by**CaptainBlack**

$\displaystyle x^7+x^6-x^4-x^3-x^2+x-1$and

$\displaystyle -x^7+x^6-x^4+x^3-x^2-x-1?$

What is the logic behind Descartes rule?

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Malay - Jul 18th 2006, 04:16 AMCaptainBlackQuote:

Originally Posted by**malaygoel**

believe that it requires all the powers less than the maximum appear with non-

zero coefficients. So it applies to:

$\displaystyle x^7+x^6+2x^5-x^4-x^3-x^2+x-1$,

where the signs are +,+,+,-,-,-,+,-, which change three times, and so this

polynomial has at most three positive roots, and so has either 1 or 3 positive

roots.

Now for the negative roots we switch the signs of the odd powers to give:

$\displaystyle -x^7+x^6-2x^5-x^4+x^3-x^2-x-1$,

now the signs are -,+,-,-,+,-,-,-, which change sign 4 times, so there

are at most four negative roots to the original polynomial. So the original

polynomial has 4, 2 or 0 negative roots.

RonL - Jul 18th 2006, 09:07 PMmalaygoelQuote:

Originally Posted by**CaptainBlack**

Malay - Jul 18th 2006, 10:25 PMCaptainBlackQuote:

Originally Posted by**Malay**

the applicability of the rule of signs to polynomials with missing powers.

(weasel words - wording which alows the author to subsequently disavow

what they wrote).

It appears that the rule of signs is applicable :o . I have looked at the

problem that I thought might exist with such polynomials again and

it turns out they are not real so we can proceed:

$\displaystyle

x^7+x^6-x^4-x^3-x^2+x-1

$

Has signature +,+,-,-,+,-. The signs chenge from + to - or - to + three

times in this signature so there are at most three positive roots.

Now to investigate the negative roots we change the signs of all the

odd power terms in the polynomial and then proceed as before:

$\displaystyle

-x^7+x^6-x^4+x^3-x^2-x-1

$

which has signature -,+-,+,-,-. The signs change 4 times in this signature,

so there are at most four negative roots.

RonL