What Are the Roots of a Given Quartic Polynomial?

In summary: There's no question mark. There's only a statement: "that's what I did...let me post my equations."I would like to see the rest of your equations, and how you got them.In summary, the conversation discusses a problem where a quartic polynomial is given along with its roots. The task is to find the value of the coefficient A. One method suggested is to make substitutions for the roots and equate coefficients to obtain four equations in four unknowns. Another approach is to substitute one of the roots for x and solve for A, but the validity of this method is questioned. The OP has solved the problem and it is no longer pending.
  • #1
chwala
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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

Homework Equations

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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  • #2
whats your question?
 
  • #3
chwala said:

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

Homework Equations

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##.
 
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  • #4
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##
 
  • #5
chwala said:
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.
 
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  • #6
chwala said:
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?
 
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  • #7
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##
 
  • #8
Any other alternative method to the quartic polynomial?
 
  • #9
Mark44 said:
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.
 
  • #10
Mark44 said:
you have four equations in four unknowns, so with some work a solution can be found.
FactChecker said:
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)##.
 
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  • #11
Mark44 said:
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.
 
  • #12
chwala said:

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 anyone 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?
 
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  • #13
kuruman said:
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 anyone 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.
 
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  • #14
kuruman said:
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 anyone 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...
 
  • #15
kuruman said:
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 anyone 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?
 
  • #16
chwala said:
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.
 
  • #17
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?
 
  • #18
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.
 
  • #19
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.
 
  • #20
chwala said:
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##?
chwala said:
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.
 
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  • #21
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##
 
  • #22
That is correct. How much of all this do you need to get ##A##?
 
  • #23
am getting;
##(abcd)^2=4##.............5
## {a^2+b^2+c^2+d^2}=1-2A##..........6
i need way forward...
 
  • #24
chwala said:
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?
 
  • #25
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?
 
  • #26
chwala said:
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?
It is not as tough as you think. Take a deep breath, clear your mind then read very carefully the following that I repeat from post #24.
kuruman said:
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##.
Can you achieve this goal by looking at equations that do not contain ##A##?
 
  • #27
Thanks for your insight, let me look at it again.
 
  • #28
Lol still getting stuck...i will post my attempts...came up with equation...
## \frac {-2} {cd} ####(c+d)+cd(a+b)##= -4......7

another attempt: simultaneous equations,
## 1/a+1/b+1/c+1/d =2##............8
## a+b+c+d = -1##

another attempt:simultaneous equation,
## 1/a^2+1/b^2+1/c^2+1/d^2 + 2/A = 4##.......9
## a^2+b^2+c^2+d^2+2A = 1##
kindly advise if i am on the right track.
 
  • #29
It seems we are talking past each other. Please explain to me, in your own words, what the problem is asking you to find. Once we agree on that, I will guide you to the next step.
 
  • #30
The problem requires that we find a numeric value for ##A## or rather the value of ##A## which is numeric ...
 
  • #31
chwala said:
The problem requires that we find a numeric value for ##A## or rather the value of ##A## which is numeric ...

Yes, but don't forget you are pretending that you know the values of the four roots, so you need a formula that, somehow, involves those roots.
 
  • #32
To illustrate what @Ray Vickson said, suppose I told you that
##\Theta = -5.50427##
##\Psi=0.08167 + 0.27755 i##
##\phi=0.08167 - 0.27755 i##
##\xi=4.34094##
Can you find a numeric value for ##A## given this set of roots? Note that there are infinitely many sets of roots because ##A## can be chosen to have infinitely many values.

On edit (2 days later)
Actually, the numbers above are the actual roots consistent with the problem's statement. In other words
##1/\Theta = -5.50427##
##1/\Psi=0.08167 + 0.27755 i##
##1/\phi=0.08167 - 0.27755 i##
##1/\xi=4.34094##

I apologize for the confusion.
 
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  • #33
let me look at it again...
 
  • #34
so reading your comments, and from my understanding, the only possibility is to use trial and error in trying to figure out the roots of the problem, this is my latest equation.
## \frac {-2} {cd}####(c+d) + cd(-1-c-d)=-4##
is the above equation correct? if so, then how do we get the values of ##c## and ##d## ?
 
  • #35
chwala said:
so reading your comments, and from my understanding, the only possibility is to use trial and error in trying to figure out the roots of the problem, this is my latest equation.
## \frac {-2} {cd}####(c+d) + cd(-1-c-d)=-4##
is the above equation correct? if so, then how do we get the values of ##c## and ##d## ?
What roots? I gave you the roots for one choice of ##A## in #32. Let me remind you of the definitions
Mark44 said:
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##.
So you know ## \Theta##, ##\Psi##, ##\xi## and ##\phi## and you can easily find ## a##, ##b##, ##c## and ##d## from the definitions.
Can you find ##A##?
 

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