標題:
a.maths (HKCEE1991)
發問:
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.
p+q+k= 2 pq+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
其他解答:
(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
a.maths (HKCEE1991)
發問:
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.
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最佳解答:p+q+k= 2 pq+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
其他解答:
(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
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