數學一元二次配方法

1.
4x² = 121
(2x)² - (11)² = 0
(2x - 11)(2x + 11) = 0
x = 11/2 或 x = -11/2

或：
4x² = 121
x² = 121/4
x = ±√(121/4)
x = 11/2 或 x = -11/2


2.
(2x - 5)² - 36 = 0
(2x - 5)² - 6² = 0
[(2x - 5) - 6] [(2x - 5 + 6) = 0
(2x - 11)(2x + 1) = 0
x = 11/2 或 x = -1/2

或：
(2x - 5)² - 36 = 0
(2x - 5)² = 36
2x - 5 = 6 或 2x - 5 = -6
2x = 11 或 2x = -1
x = 11/2 或 x = -1/2


3.
(x - 5)² - 36 = 0
(x - 5)² - 6² = 0
[(x - 5) - 6] [(x - 5 + 6) = 0
(x - 11)(x + 1) = 0
x = 11 或 x = -1

或：
(x - 5)² - 36 = 0
(x - 5)² = 36
x - 5 = 6 或 x - 5 =-6
x = 11 或 x = -1

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別忘了負數值的答案~

1.
4x^2=121
(2x)^2=121
2x=11
x=5.5

2.
(2x-5)^2-36=0
(2x-5)^2=36
2x-5=6
x=5.5


3.
(x-5)^2-36=0
(x-5)^2=36
x-5=6
x=11