✔ 最佳答案
1a)8x2﹣(k + 2)x + 2 = 0 has a double root
So Δ > 0
[-(k + 2)]2﹣4(8)(2) > 0
k2 + 4k + 4﹣64 > 0
k2 + 4k﹣60 > 0
(k + 10)(k﹣6) > 0
-10 < k < 6
2a)f(x) = 2x3﹣x2﹣5x﹣2 and g(x) = x3﹣4x2 + x + 6
f(-1) = 2(-1)3﹣(-1)2﹣5(-1)﹣2 = 0
g(-1) = (-1)3﹣4(-1)2 + (-1) + 6 = 0
So x + 1 is a common factor of f(x) and g(x)
b) 2x2﹣3x﹣2
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x + 1 / 2x3﹣x2﹣5x﹣2
2x3 + 2x2
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-3x2﹣5x
-3x2﹣3x
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- 2x﹣2
- 2x﹣2
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So f(x) = (x + 1)(2x2﹣3x﹣2) = (x + 1)(x﹣2)(2x + 1)
x2﹣5x + 6
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x + 1 / x3﹣4x2 + x + 6
x3 + x2
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-5x2 + x
-5x2﹣5x
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6x﹣6
6x﹣6
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So g(x) = (x + 1)(x2﹣5x + 6) = (x + 1)(x﹣2)(x﹣3)
ci)Let h(x) = a[f(x)] + b[g(x)]
So f(x) = 3x3﹣5x2﹣4x + 4 = (2a + b)x3﹣(a + 4b)x2 + (b﹣5a)x + (6b﹣2a)
2a + b = 3 ———(1)
a + 4b = 5 ———(2)
(2) X 2 : 2a + 8b = 10 ———(3)
(3)﹣(1) : 7b = 7
b = 1 ———(4)
put (4) into (1),
2a + 1 = 3
a = 1
So h(x) = f(x) + g(x)
ii)3x3﹣5x2﹣4x + 4 = 0
h(x) = 0
f(x) + g(x) = 0
(x + 1)(x﹣2)(2x + 1) + (x + 1)(x﹣2)(x﹣3) = 0
(x + 1)(x﹣2)[(2x + 1) + (x﹣3)] = 0
(x + 1)(x﹣2)(3x﹣2) = 0
x = -1 or x = 2 or x = 2/3
