Inequalities

Kyy010203

2009-10-31 7:15 am

Inequalities
6. find the value(s) of k so that f(x)=4x^2 + 4kx +6k -9 is non negative for all real values of x.
12 given X ,Y and m are real numbers satisfying the condition X+Y=X^2 + Y^2 = 2m+4.
a)find a quadratic equation with coefficients in terms of m and whose roots are X and Y.
b)hence, find the range of possible values of m

回答 (1)

用戶2053

2009-10-31 7:46 am

✔ 最佳答案

6)
f(x)=4x^2 + 4kx +6k -9 >= 0

So f(x)=0 have no real roots

△ = (4k)^2 - 4(4)(6k-9) < =0

16k^2 - 96k + 144 < =0

k^2 - 6k + 9 < =0

(k - 3)^2 < =0

k = 3

12a)

X+Y = (X+Y)^2 - 2XY = 2m+4 = 2(m+2)

So :

(2m+4)^2 - 2XY = 2m+4

2XY = (2m+4)^2 - (2m+4)

XY = (1/2)(2m+4)(2m+4 -1)

XY = (m+2)(2m+3)

The required equation is x^2 - (X+Y)x + XY = 0

i.e. x^2 - 2(m+2)x + (m+2)(2m+3) = 0

b)

Since X ,Y are real numbers :

△ =[2(m+2)]^2 - 4(m+2)(2m+3) >= 0

4(m^2 + 4m + 4) - 4(m+2)(2m+3) >= 0

m^2 + 4m + 4 - (2m^2 + 4m + 3m + 6) >= 0

- m^2 - 3m - 2 >= 0

m^2 + 3m + 2 <= 0

(m+1)(m+2) <= 0

- 1 <= m <= - 2

2009-10-31 02:05:06 補充：
The second line should be :

So f(x)=0 have no real roots or have double roots

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