谁能帮我解下列一元二次方程啊,x(x-14)=0x^2=x+564x^2-45=31x(x+8)(x+1)=-12x^+
3个回答
1.解1 x(x-14)=0 解2 x(x-14)=0
x-14=0 x=0
x=14
2.x^2 - x - 56 = 0
(X - 8)(x +7) = 0
x =8,-7
3.4x^2-45=31x
4x²-45=31x
4x²-31x-45=0
(x-9)(4x+5)=0
x1=9 x2=-5/4
4.（x+8）（x+1）=-12
X^2+9X+8+12=0
X^2+9x+20=0
(x+4)(x+5)=0
5.x^2+12x+27=0
十字相乘法
1 3
1 9
（x+3)(x+9)=0
x=-3或x=-9
6.x(5x+4)=5x+4
x(5x+4)-(5x+4)=0
(5x+4)(x-1)=0
x1=-4/5 x2=1
7.-3x^2+22x-24=0
3x^2-22x+24=0
(x-6)(3x-4)=0
x1=6 x2=4/3
8.(3x-2)(3+x)=x+14
3x^2+7x-6=x+14
3x^2+6x-20=0
x^2+2x-20/3=0
x^2+2x+1-1-20/3=0
(x+1)^2-23/3=0
(x+1)^2-69/9=0
(x+1-√69/3)(x+1+√69/3)=0
x=-1+√69/3或x=-1-√69/3
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