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Quartic polynomials

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chwala

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1. Homework Statement
Given ## x^4+x^3+Ax^2+4x-2=0## and giben that the roots are ## 1/Φ, 1/Ψ, 1/ξ ,1/φ##
find A


2. Homework Equations


3. The Attempt at a Solution
## (x-a)(x-b)(x-c)(x-d)=0## where a,b,c and d are the roots
 
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mathwonk

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whats your question?
 
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1. Homework Statement
Given ## x^4+x^3+Ax^2+4x-2=0## and giben that the roots are ## 1/Φ, 1/Ψ, 1/ξ ,1/φ##
find A


2. Homework Equations


3. The Attempt at a Solution
## (x-a)(x-b)(x-c)(x-d)=0## where a,b,c and d are the roots
Rather than work with all those Greek letters, I would make these substitutions:
Let ##a = \frac 1 \Theta, b = \frac 1 \Psi, c = \frac 1 \xi,d = \frac 1 \phi##.
Next, multiply out your equation. Then equate the coefficients of the ##x^3, x^2, x## terms and the constant term with those given in the original equation. Doing this, you should get four equations in the unknowns a, b, c, and d.

For example, one of the equations is ##(-a)(-b)(-c)(-d) = -2##, or equivalently, ##abcd = -2##.
 

chwala

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that's what i did....let me post my equations,
##abc +abd+acd+bcd = 0.5##
## ab+ad+ac+bd+bc+cd=-0.5A##
## a +b+c+d = 2##
 
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that's what i did....let me post my equations,
##abc +abd+acd+bcd = 0.5##
## ab+ad+ac+bd+bc+cd=-0.5A##
## a +b+c+d = 2##
It looks like you're on the right track, but I haven't worked the problem, so can't confirm that your equations are correct. With those three equations and the one from me, you have four equations in four unknowns, so with some work a solution can be found.
 

Ray Vickson

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that's what i did....let me post my equations,
##abc +abd+acd+bcd = 0.5##
## ab+ad+ac+bd+bc+cd=-0.5A##
## a +b+c+d = 2##
Where do all the "0.5"s come from?
 

chwala

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Ray just from a summary of my working. Anyway without boring you guys i realize that somewhere in the working one has to make use of the identity
## (a+b+c+d)^2 ≡ a^2 + b^2 +c^2 +d^2 + 2(ac+ad+bc+bd+cd+ab)## without which you can't arrive at the solution. Are there alternative methods?
## A=-1##
 

chwala

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Any other alternative method to the quartic polynomial?
 

FactChecker

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you have four equations in four unknowns, so with some work a solution can be found.
I'm not so sure. These are not linear equations.
 
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you have four equations in four unknowns, so with some work a solution can be found.
I'm not so sure. These are not linear equations.
I don't see how that makes a difference other than you can't use matrix methods to find a solution.
Here's a simple example of two nonlinear equations:
##x^2 + y^2 = 1##
##(x - 1)^2 + y^2 = 1##
These equations represent two circles of radius 1. The first is centered at the origin, and the second is centered at (1, 0). By subtracting the first equation from the second, you get ##2x = 1## or ##x = \frac 1 2##. Back-substitution into the first equation yields ##y = \pm \frac {\sqrt 3} 2##, making the intersection points ##(\frac 1 2, \frac {\sqrt 3} 2)## and ##(\frac 1 2, \frac {-\sqrt 3} 2)##.
 

FactChecker

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I don't see how that makes a difference other than you can't use matrix methods to find a solution.
Here's a simple example of two nonlinear equations:
##x^2 + y^2 = 1##
##(x - 1)^2 + y^2 = 1##
These equations represent two circles of radius 1. The first is centered at the origin, and the second is centered at (1, 0). By subtracting the first equation from the second, you get ##2x = 1## or ##x = \frac 1 2##. Back-substitution into the first equation yields ##y = \pm \frac {\sqrt 3} 2##, making the intersection points ##(\frac 1 2, \frac {\sqrt 3} 2)## and ##(\frac 1 2, \frac {-\sqrt 3} 2)##.
Ok. But I just don't think that there is a reliable theory regarding the existence of solutions to a number of nonlinear simultaneous equations. That being said, there might be something about these equations that can be used. I don't know.
 

kuruman

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1. Homework Statement
Given ## x^4+x^3+Ax^2+4x-2=0## and giben that the roots are ## 1/Φ, 1/Ψ, 1/ξ ,1/φ##
find A
I must be having a blind spot with this problem and it's driving me nuts. If the roots are given and they are truly roots, then isn't it true that substituting any one of them for ##x## will give an equation that can be easily solved to get a consistent value for A? What am I missing that the problem is asking?
 

FactChecker

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I must be having a blind spot with this problem and it's driving me nuts. If the roots are given and they are truly roots, then isn't it true that substituting any one of them for ##x## will give an equation that can be easily solved to get a consistent value for A? What am I missing that the problem is asking?
Ha! Of course! I am the blind one.
 

chwala

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I must be having a blind spot with this problem and it's driving me nuts. If the roots are given and they are truly roots, then isn't it true that substituting any one of them for ##x## will give an equation that can be easily solved to get a consistent value for A? What am I missing that the problem is asking?
maybe, you could post your attempt, and see where you're not getting it...
 

chwala

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I must be having a blind spot with this problem and it's driving me nuts. If the roots are given and they are truly roots, then isn't it true that substituting any one of them for ##x## will give an equation that can be easily solved to get a consistent value for A? What am I missing that the problem is asking?
did you manage to find the solution or you would like me to post it for you?
 

kuruman

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did you manage to find the solution or you would like me to post it for you?
It is against forum rules for me to post the solution. You are the OP, therefore you should post the solution and mark the problem as "solved". If my solution disagrees with yours, I will say so.
 

chwala

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I had already solved this problem, going through the threads, you seem not to understand, but you've confirmed that you know the solution. I don't think the question is still pending????
 

kuruman

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In post #4 you say
##ab+ad+ac+bd+bc+cd=-0.5A##
Is your solution then
##A=-2(ab+ad+ac+bd+bc+cd)=-2(\frac{1}{\Theta \Psi}+\frac{1}{\Theta \phi}+\frac{1}{\Theta \xi}+\frac{1}{\Psi \phi}+\frac{1}{\Psi \xi}+\frac{1}{\xi \phi})##?
The question is still pending until you are satisfied that you have the correct solution.
 

chwala

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I solved this in (post 7). I did not want to write the whole workings, let me check my files for this. It is solved already by me.
 

kuruman

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I solved this in (post 7).
In post #7 you say ##A=-1##. How can you get a numerical value for ##A## if you do not have numerical values for the roots ##\Theta##, ##\Psi##, ##\phi## and ##\xi##?
I did not want to write the whole workings, let me check my files for this. It is solved already by me.
Please check your files and post your solution in the form you think is right.
 

chwala

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ok, i am getting this equations, i can't see my files...i let the roots to be ;##a,b,c, d##
##(x-a)(x-b)(x-c)(x-d)= x^4+x^3+Ax^2+4x-2##
##a+b+c+d=-1##.............................................1
##ab+ac+ad+bc+bd+cd=A##..........................2
##bcd+acd+abd+abc= -4##.............................3
##abcd= -2##.....................................................4
is this step correct?
and further,
##(a+b+c+d)^2=a^2+b^2+c^2+d^2+2A##
 

kuruman

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That is correct. How much of all this do you need to get ##A##?
 

chwala

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am getting;
##(abcd)^2=4##....................................................................5
## {a^2+b^2+c^2+d^2}=1-2A##.............................................6
i need way forward...
 

kuruman

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am getting;
##(abcd)^2=4##....................................................................5
## {a^2+b^2+c^2+d^2}=1-2A##.............................................6
i need way forward...
So you have 4 equations in post #21 and 2 more in post #23 for a total of 6. Your goal is to find an expression for ##A##, meaning that you need to have ##A## alone on the left hand side and some expression on the right hand side involving ##a##, ##b##, ##c## and ##d##. Study each equation carefully, one at a time. How do you think you should proceed to achieve your goal?
 

chwala

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this is a tough one...still struggling, my latest attempt
## -(1/a +1/b+1/c+1/d)= abcd##..........attempt 1
and
##-(cd(a+b)+ab(c+d)=4##
##cd(-1-c-d)##+##\frac 2 {cd} (c+d)=4##........attempt 2, am i on the right path?
 

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