Mathematical Problem

(1) x^2 - 4x + k = 0 has two different solution
=> ∆ > 0
i.e. (-4)^2 - 4(1)(k) > 0
16 - 4k > 0
4k -16 < 0
4k < 16
k < 4
If k is an integer, its greatest value is 3.

(2) Tke the greatest value of k
i.e. k = 3
x^2 - 4x + 3 = 0
(x -1)(x - 3) = 0
x = 1 or 3
Since x^2 - 4x + k = 0 and x^2 + mx - 1 = 0
Therefore a factor of x^2 + mx -1 is (x - 1)
Since the constant term of it is -1 and coefficient of x^2 = 1
Therefore, another factor must be (x + 1)
consider (x+1)(x-1) = x^2 + mx -1
x^2 - 1 = x^2 + mx -1
Comparing coefficients, m = 0

That delta is the discriminant of the quadratic equation.

For any quadratic equation ax^2 + bx + c = 0 , the solution can be given by the formula:
(-b + square root (b^2 - 4ac) ) / 2a or (-b - square root (b^2 - 4ac) ) / 2a

delta = b^2 - 4ac, and this determine the number of solutions.
when delta = 0 , only one solution (because -b + or - square root (b^2 - 4ac) is just the same.
when delta > 0, two solutions can be found.

Hence, for the problem above, you may use delta > 0 , ie (-4)^2 - 4(1)(k) > 0
to find the answer

1)
x^2 - 4x + k = 0
delta = 16 - 4k > 0 (because two different solutions)
k < 4

2)
你話有無 greatest value of k,
k < 4, 唔等於 4,拎 3.99 定係 3.999999

如果當 k = 4, 問題就唔係 has two different solutions, 係 has real solutions

當 k = 4, 做埋佢,
x^2 - 4x + 4 = 0
(x - 2)^2 = 0
x = 2

sub it into x^2 + mx - 1 = 0,
4 + 2m - 1 = 0
m = -3/2

2007-06-19 17:40:16 補充：
有話 k 係 integer 咩? 樓上果位...