中四附加數學

若α,α2是二次方程 x2-12x+ｋ=0的根，試求ｋ－的值．
α+α2 = 12
α+α2 – 12 = 0
(α + 4)( α – 3) = 0
α = -4 或α = 3

同時
α*α2 = k
當α = -4
則
k = α3
= (-4)3
= -64
當α = 3
則
k = α3
= (3)3
= 27

用sum of roots=12
12=α+α^2
α=3or-4
product of roots=k
3*3^2=k or -4*(-4)^2=k
k=27 or k=-64

因為有2個不等實根
所以b^2 - 4ac > 0
(-12)^2 - 4(1)(k) > 0
144 - 4k > 0
144 > 4k
k < 36

α^2≧0 ∴α is a real no
∵α,α^2 is the real roots of the eqt.
∴D≧0
(-12)^2-4(1)(k)≧0
144-4k≧0
4k≦144
k≦36

α + α^2 = 12

α^2 + α - 12 = 0

(α - 3)(α + 4) = 0

α = 3 or α = -4

k = α(α^2)

= α^3

if α = 3 , k = 27

if α = -4, k = -64

sum of root = a^2+a=12
----> a^2+a-12=0
-----> (a-4)( a+3)=0

---> a=4 or -3

product of root a^2 x a = a^3 = k

so k = 4^3 =64 or (-3)^3 = -27