If there are multiple roots, try this one

http://www.neiu.edu/~zzeng/multroot.htm

or email: zzeng@neiu.edu

=====
Hi Roland,
thank you for the interesting reference.
However, by means of my own ,,computer" I found
x_1= - 8*phi/3 , x_2=x_3=...=x_9= phi/3 .
Regards, Alex.

Hi Roland,
there is only one polynomial of form
Q(X)=X^9 - (4*phi^2)X^7 +a(3)X^6+...+a(9)
having properties:
i) all roots are real ,
ii) Q(X) > 0 for all X in (phi/3,infty) .
More exactly
Q(X)=(X+ 8*phi/3)*(X-phi/3)^8= X^9 -(4*phi^2)X^7+.... .
Regards,Alex

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1 20th December 07:48 zhonggang zeng External User Posts: 1	If there are multiple roots, try this one http://www.neiu.edu/~zzeng/multroot.htm or email: zzeng@neiu.edu

2 21st December 22:25 alexandru.lupas External User Posts: 1	Finding the roots of Q(X)=X^9+... ===== Hi Roland, thank you for the interesting reference. However, by means of my own ,,computer" I found x_1= - 8*phi/3 , x_2=x_3=...=x_9= phi/3 . Regards, Alex.

3 22nd December 10:40 alexandru.lupas External User Posts: 1	Finding the roots of Q(X)=X^9+... Hi Roland, there is only one polynomial of form Q(X)=X^9 - (4phi^2)X^7 +a(3)X^6+...+a(9) having properties: i) all roots are real , ii) Q(X) > 0 for all X in (phi/3,infty) . More exactly Q(X)=(X+ 8phi/3)(X-phi/3)^8= X^9 -(4phi^2)X^7+.... . Regards,Alex