Each of the coefficients of the polynomial:
F(x) = x⁴ + P*X³ + Q*X² + R*X + S is real.

Determine Q, given that the equation F(x) =0 has precisely four non real roots, such that:
Two of the roots add up to 3+4i and the remaining two roots multiply up to 13+i

First if we know the roots W,X,Y,Z we can write the polynomial as
x^4 -(W+X+Y+Z)x^3+(WX+WY+WZ+XY+XZ+YZ)x^2-(WXY+WXZ+WYZ+XYZ)x+(WXYZ)

Now suppose:
W=a+bi
X=a-bi
Y=c+di
Z=c-di

It turns out that it doesn't matter which pair (W&Y, W&Z, X&Y, or X&Z) sums to 3+4i and its complement produces 13+i so let

W+Y=3+4i
XZ=13+i

Substituting the real and imaginary parts yields the system:
a+c=3
b+d=4
ac-bd=13
-ad-bc=1

Solving (many steps omitted) yields a and c as the real solutions to
-4a^4+24a^3-113a^2+231a -104=0
Wolfram alpha gives these as (6± √(2√(4265)-59))/4

(more omitted)
The approximate roots of F are
W=.612-1.942i
X=.612+1.942i
Y=2.388+5.942i
Z=2.388-5.942i

Returning to the very top we can recreate F.
F(x)=x^4-6x^3+51x^2-70x+170

Posted by Jer on 2016-02-07 16:12:32