Connection between roots of polynomials of degree n

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Homework Help Overview

The discussion revolves around the relationship between the roots of two polynomial equations of degree n that have the same coefficients but in different orders. Participants are exploring how the roots of these equations are connected.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to understand the relationship between the roots by considering specific cases, such as n=1 and n=2. Some suggest manipulating the equations by dividing by certain terms, while others propose expressing the polynomials in terms of a new variable.

Discussion Status

There is an ongoing exploration of different approaches to relate the roots of the two polynomial equations. Some participants have provided insights into specific cases, while others are questioning the implications of their manipulations. No consensus has been reached, but productive lines of inquiry are being pursued.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is also a sense of uncertainty regarding the best approach to take in analyzing the roots.

jaci55555
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Homework Statement


The two polynomial eqns have the same coefficients, if switched order:
a_0 x_n+ a_1 x_n-1 + a_2 x_n-2 + … + a_n-2 x_2 + a_n-1 x + a_n = 0 …….(1)
a_n x_n+ a_n-1 x_n-1 + a_n-2 x_n-2 + … + a_2 x_2 + a_1 x + a_0 = 0 …….(2)
what is the connection between the roots of the eqns?

2. The attempt at a solution

i don't know how to do this at all!
i thought maybe divide through by the first terms coeff... but that doesn't seem to help. maybe there is some kowledge I'm missing. something about how to work out the roots?
please help!
 
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what if [itex]n=1[/itex], what's the relation between the roots?

Then try [itex]n=2[/itex].
 
Don't divide by the coefficient. Divide by x^n.
 
algebrat:
at n=1 it is just the reciprocal
at n =2 it's: x_1 = (a_1 +- sqrt(a_1^2 -4a_2a_0))/2a_0
and x_2 = (a_1 +- sqrt(a_1^2 -4a_2a_0))/2a_2
so everything above the line is he same, just the a_0 and a_2 change
i'll find a formula for n =3 and try that...

dick: i tried this, but don't see where it leads,
a_0 + a_1/x + a_2/x^2 + a_3/x^3 +...+ a_n/x^n
and
a_n + a_n-1/x + a_n-2/x^2 + a_n-3/x^3 + ...+a_0/x^n
 
jaci55555 said:
a_0 + a_1/x + a_2/x^2 + a_3/x^3 +...+ a_n/x^n
and
a_n + a_n-1/x + a_n-2/x^2 + a_n-3/x^3 + ...+a_0/x^n

So here you have

[itex]f(x) = a_n + \frac{a_{n-1}}{x} + \frac{a_{n-2}}{x^2} + ...+\frac{a_0}{x^n}[/itex]

Make this a function of a new variable [itex]t=1/x[/itex]

Can you relate this function to the first one? then find how roots depend on each other?
 
Infinitum said:
Can you relate this function to the first one? then find how roots depend on each other?

:) you rock my world! are you really still in high school?
 
jaci55555 said:
:) you rock my world! are you really still in high school?

Yep, I'm still learning the ropes. Two more years to go further! :biggrin:

Glad you got it. :smile:
 

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