# Zeros

• Sep 28th 2008, 12:37 PM
juldancer
Zeros
Given a cubic equation, such as x^3-2x, how many rational and irrational zeros are there?

How do I figure this out?
• Sep 28th 2008, 12:42 PM
mikedwd
There can be as many zeros as the number of the degree

so there are 3 or less zeros

the descartes rule of signs states that if you count the sign changes of the function, the number of sign changes (or minus two of that number) is the number of positive roots (or roots at 0)

so x cubed - 2x has 1 sign change, so there must be 1 positive zero (if there are 3 sign changes, however, there could be 3 OR 1 positive zero)

take f(-x) and count the sign changes to find out the number of negative zeros

it becomes -x cubed + 2x and there is 1 sign change again, so there must be one negative zero

because there must be 3 zeros total, and we know there is 1 positive and 1 negative zero, we also know that there is one complex zero

EDIT oh you said rational and irrational (not complex) woops i wasn't paying attention might not have answered your question

IN THAT CASE:

we have x cubed - 2x
factor out an x
x (x squared - 2)
x=0 is a rational root
x squared=2
absolute value of x = sq. root of 2
x = positive or negative sq. root of 2
so there is a root at 0, sq. root of 2, and negative sq. root of 2

so 2 irrational and 1 rational roots

wait a minute where is the complex root? can anyone answer my question now? i've never had this happen before...
• Sep 28th 2008, 12:47 PM
Moo
Hello,
Quote:

Originally Posted by juldancer
Given a cubic equation, such as x^3-2x, how many rational and irrational zeros are there?

How do I figure this out?

$\displaystyle x^3-2x=x(x^2-2)$

use the difference of 2 squares : $\displaystyle x^2-2=(x-\sqrt{2})(x+\sqrt{2})$

Hence $\displaystyle x^3-2x=x(x-\sqrt{2})(x+\sqrt{2})$

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In order to know how many there are, you can use Descartes' rule of signs : http://www.purplemath.com/modules/drofsign.htm
• Sep 28th 2008, 12:49 PM
CaptainBlack
Quote:

Originally Posted by juldancer
Given a cubic equation, such as x^3-2x, how many rational and irrational zeros are there?

How do I figure this out?

$\displaystyle x^3-2x=x(x-\sqrt{2})(z+\sqrt{2})$

So now you tell us how many rational and irrational roots there are.

RonL
• Sep 28th 2008, 01:00 PM
mikedwd
Can anyone answer my error? I factored correctly and I've used DesCartes rule of Signs perfectly billions of times before...maybe I've just forgotten how to do it correctly but I can't find where I made my error
• Sep 28th 2008, 01:06 PM
Moo
Hi mikedwd,

Quote:

and we know there is 1 positive and 1 negative zero, we also know that there is one complex zero
This is not possible. If the coefficients of a polynomial are real, then if there is a complex zero, there is another complex zero, namely its conjugate.

The mistake is, in my opinion, in the fact that you count 0 as +0 and not -0. but I'm sorry, I don't know this rule enough to be able to spot more precisely the error...
• Sep 28th 2008, 01:07 PM
CaptainBlack
Quote:

Originally Posted by mikedwd
Can anyone answer my error? I factored correctly and I've used DesCartes rule of Signs perfectly billions of times before...maybe I've just forgotten how to do it correctly but I can't find where I made my error

There is one change of sign and one positive root. No problem.

(a zero root added to a polynomial does not alter the number of times that the signs change, and so is undetectable by the rule of signs)

RonL
• Sep 28th 2008, 01:24 PM
mikedwd
hmm I'm sure I was told that a zero root is included with positive roots in the rule of signs, but I suppose that was wrong

anyway yes I just realized (im a bit slow today apparently) that you cannot have 1 complex zero...