(1) x₁ + x₂ = (4k - 7)/9
x₁x₂ = -6k²/9 = -2k²/3
(i) x₁ = 3x₂/2
x₁x₂ = 3x₂²/2 = -2k²/3,無解
(ii) x₁ = -3x₂/2
x₁x₂ = -3x₂²/2 = -2k²/3
x₂²= 4k²/9
x₂ = ±2k/3 (a)
x₁ + x₂ = (4k - 7)/9 = -3x₂/2 + x₂
x₂ = 2(7 - 4k)/9 (b)
由(a)(b):k = 1或k = 7
(2)
x₁ + x₂ = m - 2 = 0
m = 2
此時方程變?yōu)?x² + 1 = 0,無實數(shù)解,題有問題.如果把第二題中的兩個根換成互為倒數(shù)，那能解嗎？互為倒數(shù): x₁x₂ = 1/(m² - 2) = 1m = ±√3(1) m = √3x² + (2 - √3)x + 1 = 0∆ = 3 - 4√3 < 0, 無實數(shù)解(2) m = -√3x² + (2 + √3)x + 1 = 0∆ = 3 + 4√3 > 0, 有解