inequality

2012-08-16 4:56 am

1. If 4+x^2-5kx>0 for any real number x and k is a constant, find the range of values of k.

ANS: -4/5 <k<4/5

2. Let f(x)=x^2-8x+k and g(x)=k(x+2), where k is a constant. If f(x)>g(x) for any value of x, find the range of values of k

ANS: -16<k<-4

更新1:

我想問點解 1 x^2 - 5kx + 4 > 0 但是Discriminant < 0 ,而不是>0 thank you very much for helping me

更新2:

但係我仲係唔明點解係no real root, If 4+x^2-5kx>0 for any real number x , 佢既意思係唔係話無論x係咩real number條inequalty 大過0 , 但係點解係no real root 既? 唔該哂你地:)

更新3:

我明白了,THX you very very much!!!!!!!!!!

回答 (2)

myisland8132

2012-08-16 6:36 am

✔ 最佳答案

1 x^2 - 5kx + 4 > 0

Discriminant < 0

(-5k)^2 - 4 * 4 < 0

25k^2 - 16 < 0

(k + 4/5)(k - 4/5) < 0

-4/5 < k < 4/5

2 f(x) - g(x) > 0

x^2 - 8x + k - k(x + 2) > 0

x^2 - (k + 8)x - k > 0

Discriminant < 0

(k + 8)^2 + 4k < 0

k^2 + 20k + 64 < 0

(k + 4)(k + 16) < 0

-16 < k < -4

2012-08-16 09:31:39 補充：
因為判別式大於0表示有兩根

👍 0 👎 0

小新

2012-08-16 8:23 am

D< 0 , no roots for the quadratic equation
D = 0 , there is only one root ( or two repeated roots)
D>0 , there are two distinct roots

2012-08-16 14:36:23 補充：
如果成條quadratic curve 既 y value 都大過0
即係話條curve唔會店到 x axis
另一個說法就係呢條curve 沒 real roots

👍 0 👎 0

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