Crafting an equation using roots of another equation

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Homework Help Overview

The discussion revolves around finding a polynomial equation whose roots are derived by adding 2 to the roots of a given quartic equation, specifically x^4 + 3x^3 - 13x^2 - 51x - 36 = 0. Participants are exploring the relationships between the coefficients of the original polynomial and the new polynomial formed by this transformation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to derive the coefficients of the new polynomial based on the transformation of the roots. There are discussions about the sums of the roots and products of the roots, as well as the implications of adding 2 to each root. Some participants express confusion regarding their calculations and seek clarification on where their reasoning may have gone awry.

Discussion Status

There is ongoing exploration of the problem with participants questioning their assumptions and calculations. Some hints have been provided regarding the relationship between the roots and the transformation, but there is no consensus on the correct approach or final outcome yet.

Contextual Notes

Participants note that they are not allowed to solve the original equation directly as part of their working, which adds a layer of complexity to their attempts to derive the new polynomial.

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Homework Statement


Find the equation whose roots are given by adding 2 to the roots of the equation
x^4+3x^3-13x^2-51x-36=0

Homework Equations


x^4-(Σa)x³+(Σab)x²-(Σabc)x+abcd = 0

The Attempt at a Solution



And the coefficient of x^3 to be -5
second coefficient = b/a = -Σa = 3
Since we are adding two to every root so b/a = -(Σa +2) = -(Σa +Σ2) = -(-3+8) = -5 thus the equation should have x^4-5x³

And I am stuck at x^2
=Σ(a+2)(b+2)
=Σab+2a+2b+4
=Σab+ 4Σa +16 which amounts to -9
Correct coefficient is -7

Note:- We aren't allowed to solve the given equation in any part of the working
 
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Faiq said:

Homework Statement


Find the equation whose roots are given by adding 2 to the roots of the equation
x^4+3x^3-13x^2-51x-36=0

Homework Equations


x^4-(Σa)x³+(Σab)x²-(Σabc)x+abcd = 0

The Attempt at a Solution



And the coefficient of x^3 to be -5
second coefficient = b/a = -Σa = 3
Since we are adding two to every root so b/a = -(Σa +2) = -(Σa +Σ2) = -(-3+8) = -5 thus the equation should have x^4-5x³

And I am stuck at x^2
=Σ(a+2)(b+2)
=Σab+2a+2b+4
=Σab+ 4Σa +16 which amounts to -9
Correct coefficient is -7

Note:- We aren't allowed to solve the given equation in any part of the working

Hint: If ##r## is a root of ##f(x)## what would be a root of ##f(x-2)##?
 
Faiq said:

Homework Statement


Find the equation whose roots are given by adding 2 to the roots of the equation
x^4+3x^3-13x^2-51x-36=0

Homework Equations


x^4-(Σa)x³+(Σab)x²-(Σabc)x+abcd = 0

The Attempt at a Solution



And the coefficient of x^3 to be -5
second coefficient = b/a = -Σa = 3
Since we are adding two to every root so b/a = -(Σa +2) = -(Σa +Σ2) = -(-3+8) = -5 thus the equation should have x^4-5x³+...

And I am stuck at x^2
=Σ(a+2)(b+2)
=Σab+2a+2b+4
=Σab+ 4Σa +16 which amounts to -9
It is wrong. Show your working.
The correct coefficient is really -7.

Faiq said:
Correct coefficient is -7

Note:- We aren't allowed to solve the given equation in any part of the working
 
LCKurtz said:
Hint: If ##r## is a root of ##f(x)## what would be a root of ##f(x-2)##?
r+2
 
Ehild, I am aware the coefficient is wrong, the working I have given is my wrong working and I want to know where I went wrong
 
Faiq said:
r+2
Right. Can you see how to use that to make a substitution for x in the polynomial?
 
haruspex said:
Right. Can you see how to use that to make a substitution for x in the polynomial?
I am aware of the method you are implying, but the question forbids me to use that method.
 
Faiq said:
I am aware of the method you are implying, but the question forbids me to use that method.
No, it says
Faiq said:
We aren't allowed to solve the given equation in any part of the working
which is not what I have in mind.
 
Faiq said:
Ehild, I am aware the coefficient is wrong, the working I have given is my wrong working and I want to know where I went wrong
Your notations are not clear. What do you mean with Σa?
If x1, x2, x3 x4 are the roots of the original equation, the coefficient of the x2term is
Σ(xi+2)(xj+2) for all pairs i<j: it is 6 terms as you have 6 pairs from 4 elements.
So it is (x1x2+x1x3+x1x4+x2x3+x2x4+x3x4)+2(x1+x2+x1+x3+x1+x4+x2+x3+x2+x4+x3+x4)+6*4, as you have 6 times 2*2.
Also you have 3Σxi in the parentheses of the second term, instead of 2.
 
Last edited:

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