Question

设p、q是实数,证明:方程x^2+p|x|=qx-1有4个实根的充要条件是p+|q|+2＜0

Answer 1

x>0 时

x^2+px-qx+1=0,x^2+(p-q)x+1=0

设两根为x1,x2 ,x1+x2>0,-p+q>0,p-q0,

p-q2

所以p-q