# Quartic polynomials

#### chwala

Gold Member
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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#### Mark44

Mentor
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

Gold Member
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

Mentor
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

Homework Helper
Dearly Missed
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

Gold Member
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

Gold Member
Any other alternative method to the quartic polynomial?

#### FactChecker

Gold Member
2018 Award
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

Mentor
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

Gold Member
2018 Award
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 Helper
Gold Member
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

Gold Member
2018 Award
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

Gold Member
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

Gold Member
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

Homework Helper
Gold Member
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

Gold Member
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

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

#### chwala

Gold Member
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

Homework Helper
Gold Member
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.

#### chwala

Gold Member
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

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

#### chwala

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

#### kuruman

Homework Helper
Gold Member
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

Gold Member
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?

"Quartic polynomials"

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