1. A and B are the roots of the equation x^2-(k+1)x+(k-1)=0, where k is a real number.
(a) Show that A and B are real and distinct for any real number k.
(b) If A-2=|B-2|, find the value of k.
A. maths (urgent)
2007-10-29 3:33 am
回答 (2)
2007-10-29 3:44 am
✔ 最佳答案
(a)dela(判別式)=[-(k+1)]^2-4(k-1)=k^2-2k+5
=k^2-2k+1+4
=(k-1)^2+4
>0
real and distinct for any real number k in formular
(b)A-2=|B-2|
A-2=B-2 or 2-A=B-2
A=B (.NEJ) or A+B=4
A+B=k+1
k+1=4
k=3
2007-10-29 3:46 am
(a) x^2-(k+1)x+(k-1)=0
Discriminant = (k+1)^2 - 4(1)(k-1)
= (k^2 + 2k + 1) - 4k + 4
= k^2 - 2k + 5
= (k^2 - 2k + 1) + 4
= (k-1)^2 + 4
>0 for all real numbers k
Therefore A and B are real and distinct for any real number k.
(b) A-2 = |B-2|
Then, A- 2 = B - 2 or A-2 = -(B-2)
A= B or A+ B = 4
When A = B, Sum of roots = A+ B = 2A = k+1
A = (k+1)/2
Product of roots = AB= A^2 = k-1
(k+1)^2 / 4 = k-1
k^2 + 2k + 1 = 4k-4
k^2 - 2k + 5 = 0
There is no solution.
When A+B = 4
Sum of roots = k+1 = 4
k = 3
Discriminant = (k+1)^2 - 4(1)(k-1)
= (k^2 + 2k + 1) - 4k + 4
= k^2 - 2k + 5
= (k^2 - 2k + 1) + 4
= (k-1)^2 + 4
>0 for all real numbers k
Therefore A and B are real and distinct for any real number k.
(b) A-2 = |B-2|
Then, A- 2 = B - 2 or A-2 = -(B-2)
A= B or A+ B = 4
When A = B, Sum of roots = A+ B = 2A = k+1
A = (k+1)/2
Product of roots = AB= A^2 = k-1
(k+1)^2 / 4 = k-1
k^2 + 2k + 1 = 4k-4
k^2 - 2k + 5 = 0
There is no solution.
When A+B = 4
Sum of roots = k+1 = 4
k = 3
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