1.if A and A^2 are the roots of the quadratic eq.x^2-12x+k=0,find the values of k .
2.given that A and B are the roots of the quadratic eq.x^2-3x+p=0.
If A^4+B^4=17,find the values of p.
3. the min value of the function F(x)=2x^2+3kx+4k is 7/2.find the values of k.
4. show that F(x)=1/(2x^2+6x+8) is always positve,and then find the max value of F(x).
A maths 問題
2007-06-27 5:35 am
回答 (2)
2007-06-27 6:01 am
✔ 最佳答案
(1) A + A^2 = 12 --- (1)A^2 + A - 12 = 0
( A + 4 )( A - 3 ) = 0
A = -4 or A = 3 --- (1)
A( A^2 ) = k
A^3 = k --- (2)
Put (1) in to (2).
For A = -4,
( - 4 )^3 = k
k = -64
For A = 3,
( 3 )^3 = k
k = 27
Therefore k = -64 or k = 27.
(2) A + B = 3 --- (1)
AB = p --- (2)
For A = -4,
-4 + B = 3
B = 7
Therefore
p = ( 7 )( - 4 ) = -28
For A = 3,
3 + B = 3
B = 0
Therefore
p = ( 3 )( 0 ) = 0
So the values of p is -28 or 0.
(3) F(x)=2x^2+3kx+4k
F(x) = 2 ( x^2 + 3k / 2 x ) + 4k
= 2 ( x + 3k / 4 )^2 + 4k - 2 ( 9k^2 / 16 )
= 2 ( x + 3k / 4 )^2 + 4k - 9k^2 / 8
Therefore
4k - 9k^2 / 8 = 7 / 2
32k - 9k^2 - 28 = 0
9k^2 - 32k + 28 = 0
( 9k - 14 )( k - 2 ) = 0
k = 14 / 9 or k = 2
(4) By completing square,
F(x)=1/(2x^2+6x+8)
= 1 / {2 ( x + 3/2 ) ^2 + 7 / 2}
Since 2 ( x + 3/2 ) ^2 ≥ 0 , 2 ( x + 3/2 ) ^2 + 7 / 2 > 0
Therefore F( x ) is always +VE.
The max. value = 1 / ( 7/2 ) = 2 / 7
參考: My Maths Knowledge
2007-06-27 6:12 am
1)
sum of roots = A + A^2 = 12
product of roots = A^3 = k
k = 27
2)
A + B = 3, AB = p
(A^2+B^2)^2 - 2 (A^2) (B^2) = 17
((A+B)^2 - 2 AB)^2 - 2 (A^2) (B^2) =17
(9 - 2p)^2 - 2p^2 = 17
81 - 36 p + 4p^2 +-2p^2 =17
32 - 18 p + p^2 = 0
p = 16 or 2
sum of roots = A + A^2 = 12
product of roots = A^3 = k
k = 27
2)
A + B = 3, AB = p
(A^2+B^2)^2 - 2 (A^2) (B^2) = 17
((A+B)^2 - 2 AB)^2 - 2 (A^2) (B^2) =17
(9 - 2p)^2 - 2p^2 = 17
81 - 36 p + 4p^2 +-2p^2 =17
32 - 18 p + p^2 = 0
p = 16 or 2
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