p,q and k are real numbers satisfying the following conditions:
p+q+k= 2
pq+qk+kp= 1
(a) Express pq in terms of k.
(b) Find a quadratic equation, with coefficients in terms of k,
whose roots are p and q.
Hence, find the range of possible values of k.
a.maths (HKCEE1991)
2007-07-25 5:32 pm
回答 (2)
2007-07-25 5:47 pm
✔ 最佳答案
p+q+k= 2pq+qk+kp= 1
a)
p+q= 2 - k
pq+k(q+p)= 1
pq = 1 - k(p+q) = 1 - k (2-k) =1-2k+ 2k^2
b)
(x-p)(x-q)=0
x^2-(p+q)x+pq=0
x^2 -(2-k)x + (1-2k+ 2k^2)=0
For p,q are real
delta > 0
(2-k)^2-4(1-2k+ 2k^2)>0
4-4k+k^2-4+8k-8k^2>0
4k-7k^2>0
k(4-7k)>0
0<4/7
2007-07-25 09:47:27 補充：
0 < k < 4/7
2007-07-25 11:15 pm
(a)p+q+k=2
p+q=2-k
pq+qk+kp= 1
pq+(p+q)k=1
pq=1-(p+q)k
=1-(2-k)k
=k^2-2k+1
=(k-1)^2
(b)Sum of root=p+q=2-k
Product of root=pq=(k-1)^2
∴The required quadratic equation=x^2-(2-k)+(k-1)^2=0
Discriminant=(2-k)^2-4(k-1)^2>0
k^2-4k+4-4(k^2-2k+1)>0
3k^2-4k<0
0<4/3
2007-07-25 15:16:06 補充：
It should be 0<4/3
2007-07-25 15:16:59 補充：
It should be 0 < k <4/3
p+q=2-k
pq+qk+kp= 1
pq+(p+q)k=1
pq=1-(p+q)k
=1-(2-k)k
=k^2-2k+1
=(k-1)^2
(b)Sum of root=p+q=2-k
Product of root=pq=(k-1)^2
∴The required quadratic equation=x^2-(2-k)+(k-1)^2=0
Discriminant=(2-k)^2-4(k-1)^2>0
k^2-4k+4-4(k^2-2k+1)>0
3k^2-4k<0
0<4/3
2007-07-25 15:16:06 補充：
It should be 0<4/3
2007-07-25 15:16:59 補充：
It should be 0 < k <4/3
收錄日期: 2021-04-25 14:01:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070725000051KK00774