Find the set of real values of $p$ for which the equation $ | 2x + 3 | + | 2x - 3 | = px + 6 $ has more than two solutions . 
Options : A ) ( 4 , 0 )
B) R-{ 4 , 0 , -4 } 
C) {0} 
D) None 
How to find p = 0 , without having to analyse the graph ? 

let $p \in \mathbb R$, $2x-3 < 2x+3$, so if $0\leq2x-3$, $x \in [\frac{3}{2},\infty)$, the equation turns to be:
$$4x=px+6$$
this is linear equation so there is at most one solution.
if $2x+3 \leq 0 $,$x \in (-\infty,-\frac{3}{2}]$ the equation turns to be:
$$-4x=px+6$$
this is linear equation so there is at most one solution.
if $2x-3<0$ and $0 <2x+3$, $x \in [-\frac{3}{2},\frac{3}{2}]$ we have:
$$ px=0$$
so we need $p$ such that we will have at least three solution, $0$ is always a solution,and if $p=0$, we have that the last equation has many solution(more than 3 , $x \in (-1.5,1.5)$ is a solution), so $p=0$ has more than two solutions,if $ p\neq 0$ and we have that $x=\frac{6}{(4-p)}$ and $x \in [\frac{3}{2},\infty)$ is a solution (there is one solution if $0 \leq p < 4$) and $x=-\frac{6}{(4+p)}$ and $x \in (-\infty,-\frac{3}{2}]$ (there is one solution if $-4<p\leq0$) when $p \neq 4,-4$, so we have that the set of real values $p$ that the equation has more than two solutions is $\{0\}$, the set of real valued $p$ that have two or more solutions is $(4,-4)$.
