# Quartic polynomials

Homework Helper
Gold Member
Bingo again!

#### chwala

Gold Member
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

Homework Helper
Gold Member
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

Gold Member
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, lets see how it goes, thanks though...

Homework Helper
Gold Member

#### 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

Homework Helper
Gold Member
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

Gold Member
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

Gold Member
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

Homework Helper
Gold Member
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.

Last edited:

#### chwala

Gold Member
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.

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.
I understand your point, very clear, remember you are assigning values for either $A$ or a root value. My question is, in that problem "are we able to solve without assigning the values like the way you have done?"The approach of assigning values implies that $A$ may take any value dependant on the roots. What exactly do you mean by saying that i am half way to getting the solution?Were you envisaging assigning of values?If this is the case, how would this help in solving that problem?

#### kuruman

Homework Helper
Gold Member
I understand your point, very clear, remember you are assigning values for either $A$ or a root value. My question is, in that problem "are we able to solve without assigning the values like the way you have done?"The approach of assigning values implies that $A$ may take any value dependant on the roots.
Good. You came to that understanding at post #50. This understanding is crucial to answering the question.
What exactly do you mean by saying that i am half way to getting the solution?
I mean that you reached the crucial understanding I mentioned above. I believe that without it you would not be able to each the answer and you would be going around in circles producing more and more equations.
Were you envisaging assigning of values?
Assigning values to what?
If this is the case, how would this help in solving that problem?
The problem gave you the roots $1/\Theta,~1/\Psi,~1/\phi,$ and $~1/\xi~$. If you do what I asked you in post #60, namely
"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 you will see for yourself how to get $A$ for two numerical values of the roots. From that, you should be able to deduce how to get $A$ for any set of roots such as $1/\Theta,~1/\Psi,~1/\phi,$ and $~1/\xi~$.

If you do not do what I asked you in post #60, then I will stop posting on this thread until you do.

#### chwala

Gold Member

"Quartic polynomials"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving