Question

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^+

Answer 1

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