What Are the Roots of a Given Quartic Polynomial?

[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?

 

[kuruman]

Bingo again!

 

[chwala]

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.

can you now post your solution ...

 

[kuruman]

can you now post your solution ...

I cannot because that is against forum rules and I might get in trouble with the mentors who know and see all. However I can lead you to the solution, which is exactly what this forum's attitude is towards help with solutions. You are half way to the solution after your realization in post #50. If I told you that I know (never mind how) that a solution to $x^4+x^3+Ax+4x-2=0$ is $a=−5.50427$, what would you get if you replaced $x$ with $−5.50427$ in the equation? Please write it down and post it.

 

[chwala]

I assume you are plugging in values to satisfy the quartic equation by trial and error, right? Using the 4 Vieta's equation, there is no way you can come up with a value of the root of the equations, right? I too can use trial and error to get values satisfying the equation. I will go ahead and follow your instructions, let's see how it goes, thanks though...

 

[kuruman]

I will go ahead and follow your instructions, let's see how it goes, thanks though...

Please do.

 

[PAllen]

I cannot because that is against forum rules and I might get in trouble with the mentors who know and see all. However I can lead you to the solution, which is exactly what this forum's attitude is towards help with solutions. You are half way to the solution after your realization in post #50. If I told you that I know (never mind how) that a solution to $x^4+x^3+Ax+4x-2=0$ is $a=−5.50427$, what would you get if you replaced $x$ with $−5.50427$ in the equation? Please write it down and post it.

Hmm, any complex or real number whatsoever can be chosen as one root of this equation, giving a required corresponding value of A. What you can't do, because all except one coefficient is given, is choose more than one root arbitrarily. For example, if you want 1 to be a root, you get a very simple value for A.

 

[kuruman]

Hmm, any complex or real number whatsoever can be chosen as one root of this equation, giving a required corresponding value of A. What you can't do, because all except one coefficient is given, is choose more than one root arbitrarily. For example, if you want 1 to be a root, you get a very simple value for A.

All that is correct but I don't think OP has a clear understanding of it yet. The original question provides 4 roots in symbolic form. The number I suggested, $−5.50427$, was not arbitrarily chosen but belongs to a set of 4 numerical roots for a specific (but unknown) value of $A$ (see post #32). Looking ahead to where I am going with this, the goal is to convince OP that substitution of any of the 4 roots will yield the same equation involving $A$. This goal might be achieved by having OP substitute a second root from the set and a third and fourth if necessary. I sense that, here in particular, actual numbers will make a more compelling case to OP than symbols. Once OP's understanding is clear and is able to find the specific value of $A$ I chose, answering the original question should not be too difficult.

 

[chwala]

ok i let one of the roots be $x_{1}=-2$ and the other roots were found to be $x_{2}=0.5698, x_{3}= 0.2150+1.31i, x_{4}=0.215-1.31i$ giving $A=-1$

 

[chwala]

ok i let one of the roots be $x_{1}=-2$ and the other roots were found to be $x_{2}=0.5698, x_{3}= 0.2150+1.31i, x_{4}=0.215-1.31i$ giving $A=-1$

I still don't see how this will help me solve the original problem and i still do not understand how you came up with one root being $-5.50427$ since you're of the opinion that it was not picked arbitrarily. Can you show me how you found that particular root value?

 

[kuruman]

ok i let one of the roots be $x_{1}=-2$ and the other roots were found to be $x_{2}=0.5698, x_{3}= 0.2150+1.31i, x_{4}=0.215-1.31i$ giving $A=-1$

OK, suppose as you say $A=-1$. Then the equation becomes
$x^4+x^3-x^2+4x-2=0$
Now you claim that $x_1=-2$ is a root. This means that if I replace $x$ with $(-2)$ in the equation, the left hand side must evaluate to zero. Let's see,
$(-2)^4+(-2)^3-(-2)^2+4(-2)-2=16-8-4-2=16-14=2.$ Therefore your claim that $x_1=-2$ is a root is incorrect. The point here is that once you pick $A=-1$ all 4 roots are specified. I gave you these roots in post #46. One of them is $x_1=−2.32708.$ Let's see how that works with $A=-1.$
$(−2.32708)^4+(−2.32708)^3-(−2.32708)^2+4(−2.32708)-2=29.3254-12.6018-5.4153-9.30832-2=-0.00002\approx 0.$ Thus, to within round-off accuracy, $x_1=−2.32708$ is a root for $A=-1.$

The point here is that once you pick $A$, all 4 roots are uniquely specified. Conversely, once you pick one root, $A$ and the three remaining roots are uniquely specified.

I still don't see how this will help me solve the original problem and i still do not understand how you came up with one root being $-5.50427$ since you're of the opinion that it was not picked arbitrarily. Can you show me how you found that particular root value?

To get root $-5.50427$, first I picked a numerical value for $A$. As mentioned above, once I did that all four roots were uniquely specified. I obtained all four by using Mathematica to solve the equation and posted them in #32. To see how all this helps you solve the original problem, please do what I asked in post #53 and what you said you do in post #54 but have not done yet: Substitute one root $x_1=-5.50427$ in the equation leaving $A$ alone and see what you get. For good measure, repeat with another root, $x_2=4.34094$ for the same choice of $A$ and see what you get this time. Then we'll talk again.