What Are the Roots of a Given Quartic Polynomial?

[chwala]

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

 

[mathwonk]

whats your question?

 

[Mark44]

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, $abcd = -2$.

 

[chwala]

that's what i did...let me post my equations,
$abc +abd+acd+bcd = 0.5$
$ab+ad+ac+bd+bc+cd=-0.5A$
$a +b+c+d = 2$

 

[Mark44]

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]

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]

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]

Any other alternative method to the quartic polynomial?

 

[FactChecker]

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.

 

[Mark44]

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]

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]

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?

 

[FactChecker]

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.

 

[chwala]

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...

 

[chwala]

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?

 

[kuruman]

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]

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]

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]

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]

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]

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]

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

 

[chwala]

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

 

[kuruman]

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]

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?

 

[kuruman]

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.

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 $A$?

 

[chwala]

Thanks for your insight, let me look at it again.

 

[chwala]

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.

 

[kuruman]

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.

 

[chwala]

The problem requires that we find a numeric value for $A$ or rather the value of $A$ which is numeric ...

 

[Ray Vickson]

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.

 

[kuruman]

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.

 

[chwala]

let me look at it again...

 

[chwala]

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 $d$ ?

 

[kuruman]

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

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 $A$?

 

[PAllen]

Ok, maybe I am totally confused, but it seems to me we have 4 equations in 5 unknowns (a,b,c,d,A), so we can’t get a numeric answer for A without some lucky cancellation.

[edit: oops, the 5th equation is that A can be expressed in terms of any given root directly from the starting equation]

[edit: further oops, this last equation cannot be independent of the others, so A cannot be determined to be a specific number, as others have already said].

 

[PAllen]

So, it seems to me it is easy to find a simple expression for A in terms of the 4 roots, but it is also true that specifying a numeric value for any root is sufficient to determine A, and then determine all other roots.

 

[kuruman]

So, it seems to me it is easy to find a simple expression for A in terms of the 4 roots, but it is also true that specifying a numeric value for any root is sufficient to determine A, and then determine all other roots.

That is the gist of post #12.

 

[chwala]

What roots? I gave you the roots for one choice of $A$ in #32. Let me remind you of the definitions

So you know $\Theta$, $\Psi$, $\xi$ and $\phi$ and you can easily find $a$, $b$, $c$ and $d$ from the definitions.
Can you find $A$?

i cannot find them...how did you find your values? i am illiterate here, show me how? What is wrong with equation $34$?

 

[chwala]

further are you implying that some roots are complex numbers?

 

[chwala]

Ok, maybe I am totally confused, but it seems to me we have 4 equations in 5 unknowns (a,b,c,d,A), so we can’t get a numeric answer for A without some lucky cancellation.

[edit: oops, the 5th equation is that A can be expressed in terms of any given root directly from the starting equation]

[edit: further oops, this last equation cannot be independent of the others, so A cannot be determined to be a specific number, as others have already said].

i am getting conflicting messages here. can we get the value of $A$ algebraically? or are we using trial and error to fix values for the roots. I still don't understand how the values in post $32$ were found...i agree with you that we have 5 unknowns here and it may not be possible to get a value for $A$.

 

[chwala]

So, it seems to me it is easy to find a simple expression for A in terms of the 4 roots, but it is also true that specifying a numeric value for any root is sufficient to determine A, and then determine all other roots.

specifying has to involve two values and not specifying just a value for any root...

 

[PAllen]

specifying has to involve two values and not specifying just a value for any root...

Specifying one root value easily determines A from just the quartic equation itself, as @kuruman was the first on this thread to note. And once you have A, you can find all the other roots.

It is also true that A can be expressed as a pretty simple expression in the 4 roots.

This is not inconsistent. Given a set of valid roots, the latter expression will hold. But given 4 arbitrary choices for roots, and computing A, will have probability zero that these are actually roots of the quartic with the thus computed value of A.

 

[chwala]

Kindly see attached, a fellow African teacher in my whattsap group was able to give a way forward and a solution to the problem $A$=$-1$

 

[PAllen]

You know you must have a mistake because, as several here explained, pick any numeric value of A and there are 4 roots to P(x) over the complex numbers (counting possible multiple roots separately). The reciprocals of these must also satisfy Q(x) as you've defined it, and they will. Thus, there cannot be a unique value for A.

There is an error in your second page that leads to the false conclusion that A can be uniquely determined. See if you can find it.

 

[kuruman]

i am getting conflicting messages here. can we get the value of $A$ algebraically? or are we using trial and error to fix values for the roots. I still don't understand how the values in post $32$ were found...i agree with you that we have 5 unknowns here and it may not be possible to get a value for $A$.

I found the values in post #32 by assuming a numerical value for $A$ and then finding the roots of the quartic for that particular value of $A.$ Note that I originally posted the inverses of the roots due to some confusion on my part and I edited #32 to correct that mistake. The statement of the problem as you posted in #1 says

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

So, just as the problem requires, I gave you specific values for the inverses of the roots and asked you if you can find $A$. I did this because I sensed that you were stuck in a vicious circle with the symbols $a$, $b$, $c$ and $d$ that produced more and more equations without you realizing that they are supposed to be given (i.e. known) quantities. I had hoped that if you had specific numbers to work with, you would see what's going on here - you can still do it.

further are you implying that some roots are complex numbers?

They are for the particular value of $A$ that I chose.

Kindly see attached, a fellow African teacher in my whattsap group was able to give a way forward and a solution to the problem $A$=$-1$

This algebraic manipulation recasts the original polynomial $P(x)$ as another polynomial $Q(x)$ whose roots are the inverses of the roots of $P(x)$. You can always do that as long as $x=0$ is not a root. It is not clear to me how this leads to $A=-1$.
For $A=-1$ the roots are
$1/\Theta =-2.32708$
$1/\Psi=0.406238 - 1.22682 i$
$1/\phi=0.406238 + 1.22682 i$
$1/\xi=0.514603$
That is not given by the problem.

 

[chwala]

You know you must have a mistake because, as several here explained, pick any numeric value of A and there are 4 roots to P(x) over the complex numbers (counting possible multiple roots separately). The reciprocals of these must also satisfy Q(x) as you've defined it, and they will. Thus, there cannot be a unique value for A.

There is an error in your second page that leads to the false conclusion that A can be uniquely determined. See if you can find it.

yes i can see it...$a^-2+b^-2+c^-2+d^-2 ≠ a^2+b^2+c^2+d^2$ therefore
$4+A ≠1-2A$
...unless i am missing something.

 

[PAllen]

yes i can see it...$a^-2+b^-2+c^-2+d^-2 ≠ a^2+b^2+c^2+d^2$ therefore
$4+A ≠1-2A$
...unless i am missing something.

Bingo!

 

[chwala]

Bingo!

hahahahhahahhahahah nice one , Bingo

 

[chwala]

so to be precise,in this question we may not have a unique value for $A$ because we have not been given values for any of the two roots. If we had those values, then it would have been possible to use Vieta's theorem in finding $A$ right?