the quadratic equation

1. If a and a2 are the roots of the quadratic equation x2 - 12X + k = 0 ,find the values of k.
As a and a2 are the roots of x2 - 12X + k = 0,
Sum of roots = -(-12)/1
a + a2 = 12 .... (1)
Product of roots = k/1
a(a2) = k
k = a3 .... (2)
From (1),
a + a2 = 12
a2 + a - 12 = 0
(a + 4)(a - 3) = 0
a+4 = 0 or a-3 = 0
a = -4 or a = 3.
When a = -4,
k = (-4)3 【From (2)】
k = -64
When a = 3,
k = (3)3 【From (2)】
k = 27
So k = -64 or k =27.
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2. Show that the equation x2 - 3X + k = k2 has real roots for any real values of k.
Equation:
x2 - 3X + k = k2
x2 - 3x + k - k2 = 0
x2 - 3x + k(1-k) = 0 .... (*)
Determinant △ of (*)
= (-3)2 - 4(1)[k(1-k)]
= 9 - 4k + 4k2
= (4k2 - 4k + 1) + 8
= (2k + 1)2 + 8
> 0 【As (2k + 1)2 >= 0】
As △ > 0 for all real values of k,
So (*) has real roots for all real values of k.

2007-01-11 17:34:15 補充：
小小補充：二次方程 (quadratic equation) ax²+bx+c = 0，根和 (sum of roots) = -b/a根積 (product of roots) = c/a二次方程 (quadratic equation) ax²+bx+c = 0, 判別式 (determinant) △ = b²-4ac△ ＞ 0 表示有實根△ ＜ 0 表示沒有實根△ = 0 表示有重複根

1.
Since a and a^2 are the roots of equation, then
(x-a) (x-a^2)=0
x^2-(a+a^2)x+a^3=0

By comparing this against the above equation, we have
a+a^2=12
so, a=3

Hence, k=a^3=3^3=27