2 A MATHS QUESTIONS

用戶85377

2007-09-16 5:55 am

Show the steps clearly and explain clearly.

1. The minimum value of the function f(x) = 2x^2 + 3kx + 4k is 7/2. Find the values of k.
2(a) If k is a real number, prove that the equation 4x^2 + kx - 3 = 0 has two unequal real roots and the roots are of opposite signs.
(b) If the two roots are α and -3α, find the values of k.

回答 (1)

用戶195274

2007-09-16 6:31 am

✔ 最佳答案

1. 2x^2 + 3kx + 4k =2(x^2 + 3kx/2 + (3k/4)^2) + 4k-2(3k/4)^2
=2(x+3k/4)^2 + 4k - 9k^2/8
Minimum value = 4k - 9k^2/8 = 7/2
32k-9k^2 = 28
9k^2 - 32k + 28 = 0
(k-2)(9k-14) = 0
k = 2 or 14/9

2. Discriminant = k^2 - 4(4)(-3) = k^2 + 48 > 0
Hence two roots are real and unequal.
Also, product of roots = -3/4 <0
The two roots are of opposite signs.

(b) Products of roots = -3/4 = -3α^2
α^2=1/4
α=1/2 or -1/2
When α=1/2, sum of roots = -2α=-1 = -k/4
k = 4
When α=-1/2, sum of roots =1 = -k/4
k = -4
Hence k = 4 or -4

參考: me

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