Question

因式分解法解一元二次方程

1. (x-2)²=(2x+3)²
2. 1/4x²=x-1
3. (2x+3)²=4(2x+3)
4 . x(x-5)=6
请写明过程

Answer 1

1. (x-2)²=(2x+3)²
(x-2-2x-3)(x-2+2x+3)=0
(-x-5)(3x+1)=0
所以 x=-5 或 x=-1/3

2. 1/4x²=x-1
x^2-4x+4=0
(x-2)^2=0
得 x=2

3. (2x+3)²=4(2x+3)
(2x+3)(2x+3-4)=0

所以 x=-3/2 或 x=1/2

4 . x(x-5)=6
x^2-5x-6=0
(x-6)(x+1)=0
得 x=6 或 x=-1

Answer 2

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

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

3. (2x+3)²=4(2x+3)
(2x+3)²-4(2x+3)=0
(2x+3)(2x+3-4)=0
(2x+3)(2x-1)=0
x=-3/2,x=1/2

4 . x(x-5)=6
x²-5x-6=0
(x-6)(x+1)=0
x=6,x=-1