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Coefficient From Roots (Posted on 2016-02-07 )

Each of the coefficients of the polynomial: F(x) = x^{4} + P*X^{3} + Q*X^{2} + 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

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)]
= [X^{2} - 2aX + a^{2}
+ b^{2} ][X^{2} - 2cX + c^{2} + d^{2} ] (2)
Q is the coefficient of X^{2} in F(X), and can be found by
equating coefficients of X^{2} in (2):
Q =
a^{2} + b^{2} + c^{2} + d^{2} + 4ac
= (a + c)^{2}
+ (b + d)^{2} + 2(ac – bd)
= 3^{2} +
4^{2} + 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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