Real values of $p$ for which equation has more than two solutions

Question

The number of real values of $p$ for which the equation $|2x+3|+|2x-3|=px+6$ has more than two solution.

What I try

$\displaystyle \bullet $

If $ x\leq -\frac{3}{2}$, Then $\displaystyle -(2x+3)-(2x-3)=px+6\Longrightarrow -4x=px\Longrightarrow p=-4$

(No real solution)

$\displaystyle \bullet -\frac{3}{2}<x<\frac{3}{2}$, Then $\displaystyle (2x+3)-(2x-3)=px+6$

$\displaystyle px=0\Longrightarrow p=0,x=0$

(Only one solution)

$\displaystyle \bullet x\geq \frac{3}{2},$ Then $(2x+3)+(2x-3)=px+6\Longrightarrow 4x=px+6$

So $p\neq 4$

So for more then $2$ real solution , we have $ p\in\mathbb{R}-\{-4,4,0\}$

Can anyone please help me how I solve it graphically , Thanks

Answer 1

Here is the graphic solution.

The red plot $G$ is for function $ y=|2x+3|+|2x-3|$. So you are asking for what value of slope $p$ the straight line $L: y=px+6$ intersects with the red graph $G$. Note $L$ passes $A(0,6)$. So varies the slope $p$ means to rotate $L$ with respect to point $A$.

One solution: $p\in (-\infty, -4]\cup [4,\infty)$

Two solutions: $p\in (-4,0)\cup (0,4)$

Infinite solutions: $p=0$