x²-x+1=7用十字相乘法解答一下,需要具体过程
问题描述：
x²-x+1=7用十字相乘法解答一下,需要具体过程
答
x²-x+1=7
x^2-x-6=0
(x+2)(x-3)=0
x1=-2 x2=3
答
x²-x-6=0
(x+2)(x-3)=0
x1=-2;x2=3
答
x²-x+1=7
x^2-x-6=0
x - a
* x - b
-------------------------------
-bx + ab
+ x^2 -ax
-------------------------------
x^2-(a+b)x+ab = x^2-x-6 =(x-3)(x+2)=0 x1=3,x2=-2
a+b=1 ab=-6 a=3 b=-2
答
x²-x-6=0
(x-3)*(x+2)=0
x=3或x=-2