Question

一元二次方程提ax^2+bx+c=0(a不等于0) 则有 x1+x2=-b/a x1*x2=c/a

Answer 1

a(x-x1)(x-x2)=ax^2-a(x1+x2)x+ax1x2

又a(x-x1)(x-x2)=ax^2+bx+c

故ax^2-a(x1+x2)x+ax1x2=x^2+bx+c

经x^2的系数要相等(a=a),x的系数要相等(-a(x1+x2)=b),常数项系数要相等(ax1x2=c)

整理即得：

x1+x2=-b/a

x1x2=c/a