Question

при каких значениях параметра p уравнение |x-4|=px-2 имеет два различных корня

Answer 1

$|x-4|=px-2 \ begin{cases} px-2geq 0\ left[begin{matrix}x-4=px-2 \ x-4=2-px end{matrix}right end{cases}\ begin{cases} pgeq {2over x}\ left[begin{matrix}x(1-p)=2 \ x(p+1)=6end{matrix}right end{cases}\ begin{cases} pgeq {2over x}\ left[begin{matrix}x={2over (1-p)} \ x={6over (p+1)}end{matrix}right end{cases}\ pneq pm 1\ {2over (1-p)}neq {6over (1+p)}\ 2+2pneq 6-6p\ 8pneq 4\ pneq {1 over 2}$

$Answer: begin{cases}pgeq {2over x},\pneq pm 1,\pneq {1over 2}end{cases}$