Question

解下关于x的方程：ax²+（4a+1）x+4a+2=0（a≠0﹚ ﹙m-2﹚x²＋2mx＋m＋3=0

Answer 1

第一个方程

△=(4a+1)^2-4a(4a+2)

=(4a+1)^2-(4a+1-1)(4a+1+1)

=(4a+1)^2-(4a+1)^2+1

=1

x=[-(4a+1)±根号△]/2a=(-4a-1±1)/2a

x1=(-4a-1+1)/2a=-2

x2=(-4a-1-1)/2a=-4a/2a-2/2a=-2-1/a

第二个方程

△=(2m)^2-4(m-2)(m+3)

=4m^2-4(m^2+m-6)

=4[m^2-(m^2+m-6)]

=4(-m+6)

=-4m+24>=0

解得m