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F.4 A. Maths Question
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Consider the expression (a - 2)(2 - b): (a - 2)(2 - b) = 2a + 2b - 4 - ab = 2(a + b) - ab - 4 The equation is 2x2 - (m + 5)x + (1 + 2m) = 0 So a + b = (m + 5)/2 and ab = (1 + 2m)/2 Then 2(a + b) - ab - 4 = m + 5 - (1 + 2m)/2 - 4 = m + 1 - 1/2 - m = 1/2 > 0 So (a - 2)(2 - b) > 0 and hence there are 2 poss.: (i) a > 2 and 2 > b (ii) a < 2 and 2 < b Since a > b. (ii) will be impossible. Thus a > 2 > b
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F.4 A. Maths Question
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a and b are distinct real roots ( a > b) of equation (m + 5)x - 2x^2 - (1 + 2m) = 0, where m is a real number. Prove that a > 2 > b. (Please use F.4 A. Maths methods)最佳解答:
Consider the expression (a - 2)(2 - b): (a - 2)(2 - b) = 2a + 2b - 4 - ab = 2(a + b) - ab - 4 The equation is 2x2 - (m + 5)x + (1 + 2m) = 0 So a + b = (m + 5)/2 and ab = (1 + 2m)/2 Then 2(a + b) - ab - 4 = m + 5 - (1 + 2m)/2 - 4 = m + 1 - 1/2 - m = 1/2 > 0 So (a - 2)(2 - b) > 0 and hence there are 2 poss.: (i) a > 2 and 2 > b (ii) a < 2 and 2 < b Since a > b. (ii) will be impossible. Thus a > 2 > b
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