請教解方程急急急

For the equation ax² + bx + c = 0
x = [-b ± √(b² - 4ac)] / 2

x² - (p + q)x + (p + 1)(q + 1) = 0
x = {(p + q) ± √[(p + q)² - 4(p + 1)(q + 1)]} / 2
x = [(p + q) ± √(p² + 2pq + q² - 4pq - 4p - 4q - 4)] / 2
x = {(p + q) ± √[(p² - 2pq + q²) + (-4p - 4q) - 4]} / 2
x = {(p + q) ± √[(p - q)² - 4(p + q) - 4]} / 2

2012-08-25 23:12:15 補充：
若題目是 x² - (p + q)x + (p + 1)(q + 1) = 0
答案如上。

若題目是 x² - (p + q)x + (p + 1)(q - 1) = 0
則 [x - (p + 1)] [x - (q - 1)] = 0
x = p + 1 或 x = q - 1

2012-08-25 23:12:26 補充：
若題目是 x² - (p + q)x + (p - 1)(q + 1) = 0
則 [x - (p - 1)] [x - (q + 1)] = 0
x = p - 1 或 x = q + 1

但若題目是 x² - (p + q + 2)x + (p + 1)(q + 1) = 0
則 [x - (p + 1)] [x - (q + 1)] = 0
x = p + 1 或 x = q + 1

很明顯，樓上的答案是錯誤的。

2012-08-28 02:25:00 補充：
方法一 :
x² - (p + q)x + (p + 1)(q - 1) = 0
x = {(p + q) ± √[(p + q)² - 4(p + 1)(q - 1)]} / 2
x = [(p + q) ± √(p² + 2pq + q² - 4pq - 4p + 4q - 4)] / 2
x = {(p + q) ± √[(p² - 2pq + q²) - (4p - 4q) + 4]} / 2
x = {(p + q) ± √[(p - q)² - 2*(p - q)*(2) + (2)²]} / 2

2012-08-28 02:25:08 補充：
x = {(p + q) ± √[(p - q + 2)²]} / 2
x = [(p + q) ± (p - q + 2)] / 2
x = [(p + q) + (p - q + 2)] / 2 或 x = [(p + q) - (p - q + 2)] / 2
x = (2p + 2) / 2 或 x = (2q - 2) / 2
x = p + 1 或 x = q - 1

2012-08-28 02:25:40 補充：
方法二 :
用交叉相乘法：
x ..... -(p + 1)
x ..... -(q - 1)

[-(p + 1)] * [-(q - 1) = (p + 1)(q - 1)
[-(p + 1)] + [-(q - 1) = -p - 1 - p + 1 = -(p + q)

x² - (p + q)x + (p + 1)(q - 1) = 0
所以 [x - (p + 1)] [x - (q - 1)] = 0
x = p + 1 或 x = q - 1

x={(p+q)±√[(p+q)^2-4(1)(p+1)(q+1)]}/2
x=[(p+q)±√(p^2+2pq+q^2-4(pq+p+q+1)]/2
x=[(p+q)±√(p^2+2pq+q^2-4pq-4p-4q-4)]/2
x=[(p+q)±√(p^2-2pq+q^2-4p-4q-4)]/2
x={(p+q)±√[(p-q)^2-4(p+q+1)]}/2

If x = p+1 or q+1,
why Sum of root = p+q?