Question

用配方法解下列方程: (1)x^2-2x-5=0 (2)3x^2+2x-1=0

Answer 1

（1）解：x ² - 2 x - 5 = 0
x ² - 2 x + 1 - 1 - 5 = 0
（x - 1）² - 6 = 0
（x - 1）² = 6
x - 1 = ± √6
x = ± √6 + 1
x1 = √6 + 1
x2 = - √6 + 1

（2）解：3 x ² + 2 x - 1 = 0
x ² + （2 / 3）x - 1 / 3 = 0
x ² + （2 / 3）x + （1 / 3）² - （1 / 3）² - 1 / 3 = 0
（x + 1 / 3）² - 1 / 9 - 1 / 3 = 0
（x + 1 / 3）² = 3 / 9 + 1 / 9
（x + 1 / 3）² = 4 / 9
x + 1 / 3 = ± 2 / 3
x = ± 2 / 3 - 1 / 3
x1 = 2 / 3 - 1 / 3 = 1 / 3
x2 = - 2 / 3 - 1 / 3 = - 1

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Answer 2

x^2-2x-5=0
x^2-2x=5
x^2-2x+1=6
(x-1)^2=6
x-1=±√6
x1=√6+1
x2=-√6+1

3x^2+2x-1=0
3x^2+2x=1
x^2+2/3x=1/3
x^2+2/3x+1/9=1/3+1/9
(x+1/3)^2=4/9
x+1/3=±2/3
x1=1/3，x2=-1

Answer 3

(1)(x-3)^2=9
(2)(x+2/3)^2=7/9