✔ 最佳答案
(1)
Δ = (b² + c² - a²)² - 4b²c²
= (b² + c² - 2bc - a²) (b² + c² + 2bc - a²)
= ((b - c)² - a²) ((b + c)² - a²)
= (b - c - a)(a + b - c) (b + c - a)(a + b + c)
= - (a + c - b)(a + b - c) (b + c - a)(a + b + c)
由三角形二邊和大於第三邊得 a + c - b > 0 , a + b - c > 0 , b + c - a > 0 , 且三角形周界 a + b + c > 0,
則 △ < 0 , 判定 b²x² + (b² + c² - a²)x + c² = 0 無實根。
(2)
(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0
3x² - 2(a + b + c)x + ab + bc + ca = 0
Δ = 4(a + b + c)² - 4(3)(ab + bc + ca) = 0
2a² + 2b² + 2c² + 4ab + 4bc + 4ca - 6(ab + bc + ca) = 0
(a² - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ca + a²) = 0
(a - b)² + (b - c)² + (c - a)² = 0
a - b = b - c = c - a = 0
a = b = c
此三角形為正三角形。