What Are the Roots of a Given Quartic Polynomial?

[chwala]

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 dependent 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]

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 dependent 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]

Thank you for your insight.