標題:
Inequality
發問:
Consider a quadratic function y = 3x^2 + (2 + k)x + 3k. If the value of y is greater than 18 for all real values of x, find the range of possible values of k. Ans: 10 < k < 22
最佳解答:
So to speak: 3x2 + (2 +k)x + 3k - 18 = 0 has no real roots and so: (2 + k)2 - 4 x 3 (3k - 18) < 0 k2 + 4k + 4 - 36k + 216 < 0 k2 - 32k + 220 < 0 (k - 10)(k - 22) < 0 10 < k < 22 2011-02-10 22:16:11 補充： 加插多一句: For a curve y = 3x^2 + (2 +k)x + 3k - 18, it is opening upward and therefore, in case if the equation 3x^2 + (2 +k)x + 3k - 18 = 0 has no real roots, we can ensure that 3x^2 + (2 +k)x + 3k - 18 > 0 for all real values of x.
其他解答:
你所用的方法並不嚴謹 discriminant < 0 跟住solve inequality done~ 雖然答案有比到你, 但係你有沒有諗過10<22 可能會同range of y (y > 18) 有contradiction? 2011-02-10 22:12:07 補充： 10 < k < 22 不是 10 < 22 :P
Inequality
發問:
Consider a quadratic function y = 3x^2 + (2 + k)x + 3k. If the value of y is greater than 18 for all real values of x, find the range of possible values of k. Ans: 10 < k < 22
最佳解答:
So to speak: 3x2 + (2 +k)x + 3k - 18 = 0 has no real roots and so: (2 + k)2 - 4 x 3 (3k - 18) < 0 k2 + 4k + 4 - 36k + 216 < 0 k2 - 32k + 220 < 0 (k - 10)(k - 22) < 0 10 < k < 22 2011-02-10 22:16:11 補充： 加插多一句: For a curve y = 3x^2 + (2 +k)x + 3k - 18, it is opening upward and therefore, in case if the equation 3x^2 + (2 +k)x + 3k - 18 = 0 has no real roots, we can ensure that 3x^2 + (2 +k)x + 3k - 18 > 0 for all real values of x.
其他解答:
你所用的方法並不嚴謹 discriminant < 0 跟住solve inequality done~ 雖然答案有比到你, 但係你有沒有諗過10<22 可能會同range of y (y > 18) 有contradiction? 2011-02-10 22:12:07 補充： 10 < k < 22 不是 10 < 22 :P
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