Question

因式分解法解下列一元二次方程

⑴x²+x=0 ⑵x²-2√3 x=0 ⑶3x²-6x=-3 ⑷4x²-121=0 ⑸3x（2x+1）=4x+2 ⑹（x-4）²=（5-2x）

Answer 1

x^2+x=0
x(x+1)=0
x1=0 x2=-1

x^2-2√3x=0
x(x-2√3)=0
x1=0 x2=2√3

3x^2-6x=-3
x^2-2x=-1
x^2-2x+1=0
(x-1)^2=0
x1=x2=1

4x^2-121=0
(2x+11)(2x-11)=0
x1=-11/2 x2=11/2

3x(2x+1)=4x+2
3x(2x+1)-2(2x+1)=0
(2x+1)(3x-2)=0
x1=-1/2 x2=2/3

(x-4)^2=(5-2x)^2
(x-4)^2-(5-2x)^2=0
(x-4+5-2x)(x-4-5+2x)=0
(1-x)(3x-9)=0
x1=1 x2=3