Question

已知关于x的一元二次方程ax+bx+1=0(a不等于零）有两个相等的实数根，求ab^2/(a-2)+b^2-4

求解题过程

Answer 1

ax^2+bx+1=0(a不等于零）有两个相等的实数根
判别式 = b^2-4a = 0
b^2 = 4a
ab^2/[(a-2)^2+b^2-4] = a*(4a)/[(a-2)^2+4a-4 ] = 4a^2 / [a^2-4a+4+4a-4 ] = 4a^2 / a^2 = 4

Answer 2

由ax+bx+1=0有两相等实根可知：b^2-4a=0,即b^2=4a
ab^2/(a-2)+b^2-4
=(4a^2+4a(a-2)-4(a-2))/(a-2)
=(8a^2-12a+8)/(a-2)
=4(a-1)(a-2)/(a-2)
=4a-4

Answer 3

∵ax²+bx+1=0有两个相等的实数根
∴a≠0且
b²-4a=0
∴b²=4a
∴ab²/(a-2)+b²-4=4a²/(a-2)+4a-4