方程x2+12x+36=0,x1=?,x2=?,x1x2+=?,x1+x2=?
问题描述:
方程x2+12x+36=0,x1=?,x2=?,x1x2+=?,x1+x2=?
答
∵x²+12x+36=0 ∴(x+6)²=0
∴x1=x2=﹣6 x1+x2=﹣12 x1x2=36x²-6x-2=0(x-3)²=11∴x1=3-√11x2=3+√11 x1+x2=6 x1x2=-23x²+4x-7=0 (x-1)(3x+7)=0 ∴x1=1x3=﹣7/3x1+x2=﹣4/3x1x2=﹣7/3