## [問題]級數問題

### [問題]級數問題

[x - (a - 3d)][x - (a - d)][x - (a + d)][x - (a + 3d)] = 0 ...... (1)

(a - 3d) + (a - d) + (a + d) + (a + 3d) = 4

(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

(x + 5)(x + 1)(x - 3)(x - 7) = 0 ...... (2)

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 ■

benice

g(x) = x^4 - 40x² + (b - 76)x + (b + c - 37)。

g(x) = (x²)² - 40x² + (c + 39)。

x² = [40 ± √(40² - 4(c + 39))] / 2   (二次方程式的公式解)
x² = 20 ± √(361 - c)
x = ±√(20 ± √(361 - c))   (註：c 須滿足 -39 ＜ c ＜ 361)

-√(20 + √(361 - c)), -√(20 - √(361 - c)), √(20 - √(361 - c)), √(20 + √(361 - c))

√(20 + √(361 - c)) - √(20 - √(361 - c)) = 2 √(20 - √(361 - c))
√(20 + √(361 - c)) = 3 √(20 - √(361 - c))
20 + √(361 - c) = 9[20 - √(361 - c)]
√(361 - c) = 16
361 - c = 16²
c = 105

(圖 1-1)  y = g(x) 參數 b 動畫 (c = 105)

(圖 1-2)  y = g(x) 參數 b 動畫 (c = 200)

(圖 1-3)  y = g(x) 參數 b 動畫 (c = 25)

(圖 2)  y = g(x) 參數 c 動畫 (b = 76)

benice