A Math....

1. A + B = -k ... (1)
AB = -29 ... (2)
Since A2 + B2 = 65 (given)
(A + B)2 - 2AB = 65
k2 - 2(-29) = 65
k2 = 7
k = ±√7
2. Since A is the common root,
A2 - 4A + k = 0 ... (1)
2A2 - 3A + k = 0 ... (2)
(2) - (1): A2 + A = 0
A(A + 1) = 0
A = 0　or　A = -1
Substitute A = 0 into (1), k = 0
Substitute A = -1 into (1),
1 + 4 + k = 0
k = -5


2007-10-16 22:00:25 補充：
Checking: for k=0,x^2 - 4x ＋ 0= 0　- 　x=0 or x=42x^2 - 3x ＋ 0 = 0　x=0 or x=3/2So k=0 is correct.

2007-10-16 22:02:49 補充：
for k= -5,x^2 - 4x - 5 = 0x= -1 or x=52x^2 - 3x - 5 = 0x= -1 or x=5/2So k= -5 is also correct.

1. x2 + kx – 29 = 0

Sum of roots, A + B = -k

Product of roots, AB = -29

If A2 + B2 = 65

(A + B)2 – 2AB = 65

(-k)2 – 2(-29)0 = 65

k2 = 7

k = √7 or -√7


2. x2 – 4x + k = 0

Let A and B be the roots,

Sum of roots, A + B = 4

Product of roots, AB = k

2x2 – 3x + k = 0

Let A and C be the roots,

Sum of roots, A + C = 3/2

Product of roots, AC = k/2

From A + B = 4, we know that B = 4 – A

So, B = k/A

So, k = A(4 - A)

Since A + C = 3/2

C = 3/2 – A

So, AC = k/2

A(3/2 - A) = A(4 - A)/2

3A – 2A2 = -A2 + 4A

A2 + A = 0

A(A + 1) = 0

A = 0 (rejected) or -1

So, when A = -1, B = 5

Then k = -5