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

Let the two pairs of conjugate roots be a + ib, a – ib and
c+id, c – id.

We are given (w.l.o.g.):  a + ib  +  c + id   = 3 + 4i

and                               (a – ib)(c – id)  =  13 + i

Equating real and imaginary parts of both equations gives:

  a + c = 3,  b + d = 4,  ac – bd = 13,  ad + bc = -1          (1)

F(X)  = [X – (a + ib)][X – (a - ib)][X – (c + id)][X – (c - id)]

      = [X² - 2aX + a² + b²][X² - 2cX + c² + d²]                  (2)

Q is the coefficient of X² in F(X), and can be found by

equating coefficients of X² in (2):

                   Q   = a² + b² + c² + d² + 4ac

                        = (a + c)² + (b + d)² + 2(ac – bd)

                        = 3² + 4² + 2*13           using (1)

                    Q = 51

I’m wondering why KS didn’t ask us for the values of P, R and S?
- also integers.

Posted by Harry on 2016-02-07 16:19:05