amaths

Q1.
x^3 - 4x^2 + 9x - 10 = 0
By trial an error, put x = 2, so
(2)^3 - 4(2)^2 + 9(2) - 10 = 8 - 16 + 18 - 10 = 0,
so by Factor Theorem, x = 2 is a root of the equation and (x - 2) is a factor of the polynomial.
So x^3 - 4x^2 + 9x - 10 = (x - 2)(x^2 - 2x + 5) = 0
that means x^2 - 2x + 5 = 0
x = [2 +/- sqrt(4 - 20)]/2 = [2 +/- sqrt(-16)]/2 = (2 +/- 4i)/2 = 1 +/- 2i are the 2 complex roots.
Q2.
Let a be one of the roots, so the other root = a - 1.
so sum of roots = 2a - 1 = - b......(1) and
a(a - 1) = c...........(2)
From (1), a = (1 - b)/2, sub into (2) we get
(1- b)/2[(1 - b)/2 - 1] = c
(1 - b)(1 - b - 2) = 4c
(1 - b)(-b - 1) = 4c
1 - b^2 = - 4c
b^2 - 4c = 1.
Q2.2
Since the roots are consecutive integers, that means the roots are differ by 1, so using result of Q2.1
(4k - 5)^2 - 4k = 1
16k^2 + 25 - 40k - 4k = 1
16k^2 - 44k + 24 = 0
4k^2 - 11k + 6 = 0
(4k - 3)(k - 2) = 0
so k = 3/4 or 2.
When k = 3/4, the equation is
x^2 - 2x + 3/4 = 0
4x^2 - 8x + 3 = 0
(2x - 3)(2x - 1) = 0
the roots are x = 3/2 or 1/2.
When k = 2, the equation is
x^2 + 3x + 2 = 0
(x + 1)(x + 2) = 0
so x = - 1 or - 2.
since the roots must be integers, the final answer is -1 or - 2 (differ by 1).





2009-10-13 06:44:26 補充：
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