Question

用因式分解法解一元二次方程

4(3x-2)的平方=9(x+4)的平方

Answer 1

4(3x-2)的平方=9(x+4)的平方
(6x-4)²-(3x+12)²=0
(6x+4+3x+12)(6x+4-3x-12)=0
(9x+16)(3x-8)=0
x1=-16/9,x2=8/3

Answer 2

4(3x-2)²=9(x+4)²
4(3x-2)²-9(x+4)²=0
[2(3x-2)-3(x+4)][2(3x-2)+3(x+4)]=0
(6x-4-3x-12)(6x-4+3x+12)=0
(3x-16)(9x+8)=0
3x-16=0 9x+8=0
x=16/3或x=-8/9

Answer 3

4(3x-2)^2=9(x+4)^2
(6x-4)^2-(3x+12)^2=0
(9x+8)(3x-16)=0
x=-8/9或x=16/3

Answer 4

(6x-4)^2=(3x+12)^2
6x-4=3x+12
x=16/3