by mathslug » Thu Jan 03, 2008 1:48 am
Interesting problem. I think I got the right answer, but I REALLY need someone to check it.
Let q, r, s, and t be the roots to x^4 + ax^3 + bx^2 + cx + d = 0.
qr = (13+i)
s+t = (3+4i)
The sum of the roots taken 2 at a time = b.
b = qr + qs + qt + rs + rt + st --> b = qr + (q+r)(s+t) + st
a,b,c, and d must be real numbers. Remember that d = (qr)(st), and since it must be real, st is equal to (13-i). B must also be real, and since qr + st produces a real number the product (q+r)(s+t) must be real as well. Therefore (q+r) = (3-4i). So...
b = (13+i) + (3-4i)(3+4i) + (13-i) = 13 + 25 + 13 = 51
b = 51