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Coefficient From Roots (Posted on 2016-02-07) Difficulty: 3 of 5
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

No Solution Yet Submitted by K Sengupta    
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Solution Solution - another approach Comment 2 of 2 |
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
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