由 benice 於 星期一 十一月 14, 2016 10:05 pm
設原方程式的四根為 a - 3d, a - d, a + d, a + 3d,其中 d≧0。
則
[x - (a - 3d)][x - (a - d)][x - (a + d)][x - (a + 3d)] = 0 ...... (1)
將 (1) 展開後的 x³ 項係數與原方程式比較得
(a - 3d) + (a - d) + (a + d) + (a + 3d) = 4
所以,a = 1。
將 (1) 展開後的 x² 項係數與原方程式比較得
(a - 3d)(a - d) + (a - 3d)(a + d) + (a - 3d)(a + 3d) + (a - d)(a + d) + (a - d)(a + 3d) + (a + d)(a + 3d) = -34
(a - 3d)[(a - d) + (a + d) + (a + 3d)] + a² - d² + (a + 3d)[(a - d) + (a + d)] = -34
(a - 3d)(3a + 3d) + a² - d² + (a + 3d)(2a) = -34
3(a² - 2ad - 3d²) + a² - d² + 2a² + 6ad = -34
6a² - 10d² = -34
6 - 10d² = -34 (代入 a = 1)
10d² = 40
d² = 4
所以,d = 2。
將 a = 1, d = 2 代入 (1) 得
(x + 5)(x + 1)(x - 3)(x - 7) = 0 ...... (2)
將 (2) 展開後的 x 項係數及常數項係數與原方程式比較得
b = (5)(1)(-3) + (5)(1)(-7) + (5)(-3)(-7) + (1)(-3)(-7) = -15 - 35 + 105 + 21 = 76
c = (5)(1)(-3)(-7) = 105
c - b = 105 - 76 = 29 ■