Question

因式分解法解一元二次方程!

1.4X^2-x=0
2.(x+3)^2=9
3.2(x+1)^2+3(x+1)=0
还有,能告诉下怎么用因式分解法吗,除了a*b=0这种我好像都看不懂额

Answer 1

1.4x^2-x=0
x(4x-1)=0
x1=0,x2=1/4
2.(x+3)^2=9
(x+3)^2-9=0
(x+3）^2-3^2=0
(x+3+3)(x+3-3)=0
x(x+6)=0
x1=0,x2=-6
3.2(x+1)^2+3(x+1)=0
(x+1)[2(x+1)+3]=0
(x+1)(2x+5)=0
x1=-1,x2=-5/2

Answer 2

4X²-X=0
X（4X-1）=0
X=0 OR X=1/4

X²+6X+9-9=0
X²+6X=0
X（X+6）=0
X=0 OR X=-6

2（X²+2X+1）+3X+3=0
2X²+4X+2+3X+3=0
2X²+7X+5=0
△=b²-4ac
=49-4*5*2
=29
X=（-b±根号△）÷2
=(-7±根号29)÷2

X=2分之-7+根号29 OR X=2分之-7-根号29

Answer 3

1.4X
2.0
3.X1=-1 X2=2分之3