✔ 最佳答案
1)2kx^2 - (3k+2)x + (k+1) = 0
設 α , β 為方程式的根
則 α+β = - [ -(3k+2) ] / 2k = (3k+2) / 2k
αβ = (k+1) / 2k
a)二根之和為2
α+β = 2
(3k+2) / 2k = 2
3k+2 = 4k
2 = 4k - 3k
k = 2 //
b)二根同值異號
α+β = 0 (因為 α = -β)
(3k+2) / 2k = 0
3k+2 = 0
3k = -2
k = -2/3 //
c)二根互為倒數
αβ = 1 (因為 α = 1/β)
(k+1) / 2k = 1
k+1 = 2k
1 = 2k - k
k = 1
d)一根為零
代 x=0 入方程式,得 k+1 = 0
k = -1 //
2) x^2 - (m-2)x - (m+3) = 0
設 α , β 為方程式的根
則 α+β = - [ -(m-2) ] / 1 = m-2
αβ = -(m+3) / 1 = -(m+3)
二根之平方和為25
則 α^2 + β^2 = 25
(α+β)^2 - 2αβ = 25
(m-2)^2 - 2[-(m+3)] = 25
m^2 - 4m + 4 +2m + 6 - 25 = 0
m^2 - 2m - 10 = 0
(m+3)(m-5) = 0
m = -3 或 m = 5