# Roots of equation

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• Jul 16th 2006, 09:16 PM
malaygoel
Roots of equation
How many negative roots has the equation
$x^6 - bx^5 - 2ax^3 - cx + a^2=0, b\geq0, c\geq0?$

Keep Smiling
Malay
• Jul 17th 2006, 11:00 AM
galactus
Since $f(-x)=x^{6}+bx^{5}+2ax^{3}+cx+a^{2}$ 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, 11:50 AM
Quick
Quote:

Originally Posted by galactus
Since $f(-x)=x^{6}+bx^{5}+2ax^{3}+cx+a^{2}$ 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?.

It would be...
$f(-x)=-x^{6}+bx^{5}+2ax^{3}+cx+a^{2}$
which does have a variation.
• Jul 17th 2006, 03:17 PM
CaptainBlack
Quote:

Originally Posted by galactus
Since $f(-x)=x^{6}+bx^{5}+2ax^{3}+cx+a^{2}$ 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?.

I thought for Descartes rule to be applicable the coefficients of all the
powers up to the maximum must be non-zero.

RonL
• Jul 17th 2006, 03:51 PM
galactus
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 $(-x)^{6}+2a(-x)^{3}+a^{2}=x^{6}-2ax^{3}+a^{2}$

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

Originally Posted by galactus
BTW, Quick, $(-x)^{6}=x^{6}$. May I ask where you got the negative from?. Maybe I am missing something.

I assumed that since you changed all the negatives to positive than you would change the positives to negative, but I forgot that even-numbered exponents are automatically positive :o
• Jul 17th 2006, 08:38 PM
malaygoel
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, 11:54 PM
CaptainBlack
Quote:

Originally Posted by malaygoel
I am not familiar with Descartes rule. Please guide me.

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

Keep Smiling
Malay

Thw Mathworld link is here

RonL
• Jul 18th 2006, 03:26 AM
malaygoel
Quote:

Originally Posted by CaptainBlack
Thw Mathworld link is here

RonL

What is meant by sign changes?How will you calculate sign changes in $x^7+x^6-x^4-x^3-x^2+x-1$and $-x^7+x^6-x^4+x^3-x^2-x-1?$
What is the logic behind Descartes rule?

Keep Smiling
Malay
• Jul 18th 2006, 03:27 AM
malaygoel
Quote:

Originally Posted by CaptainBlack
Thw Mathworld link is here

RonL

What is meant by sign changes?How will you calculate sign changes in
$x^7+x^6-x^4-x^3-x^2+x-1$and
$-x^7+x^6-x^4+x^3-x^2-x-1?$
What is the logic behind Descartes rule?

Keep Smiling
Malay
• Jul 18th 2006, 05:16 AM
CaptainBlack
Quote:

Originally Posted by malaygoel
What is meant by sign changes?How will you calculate sign changes in
$x^7+x^6-x^4-x^3-x^2+x-1$and
$-x^7+x^6-x^4+x^3-x^2-x-1?$
What is the logic behind Descartes rule?

Keep Smiling
Malay

First I believe Descartes rule does not work for these polynomials. To work I
believe that it requires all the powers less than the maximum appear with non-
zero coefficients. So it applies to:

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

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:

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

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, 10:07 PM
malaygoel
Quote:

Originally Posted by CaptainBlack
First I believe Descartes rule does not work for these polynomials. To work I
believe that it requires all the powers less than the maximum appear with non-
zero coefficients. So it applies to:

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

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:

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

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

The expressions were given in your link

Malay
• Jul 18th 2006, 11:25 PM
CaptainBlack
Quote:

Originally Posted by Malay
Quote:

Originally Posted by CaptainBlack
First I believe Descartes rule does not work for these polynomials. To work I
believe that it requires all the powers less than the maximum appear with non-
zero coefficients.

RonL

The expressions were given in your link

Malay

You will note the what we call weasel words in my earlier post about
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:

$
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:

$
-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