Find values of the parameter m for absolute value equation to have two positive solutions

Question

I solved an equation with absolute value and parameter and I got two different results when I applied two methods for this question. Could you please help me to spot and explain what am I doing wrong?

Question: Find values of the parameter $m$, for which the equation $\quad x−|4−2x|=2m \quad $ has 2 positive solutions.

My first attempt was based on the definition of the absolute value:

$$|x| = \begin{cases} x, & \text{if } x \geq 0, \\ -x, & \text{if } x < 0. \end{cases} $$

so, I got:

$$|4-2x| = \begin{cases} 2x-4, & \text{if } x \geq 2, \\ 4-2x, & \text{if } x < 2. \end{cases} $$

Then, in the calculated domains, I found the x value. I had:

$$1^\circ \quad x + 4 - 2x = 2m \implies x = 4-2m \quad\text{if}\quad x \geq 2$$ $$2^\circ \quad x - 4 + 2x = 2m \implies x = \frac{4+2m}{3} \quad\text{if}\quad x < 2$$

Next, I thought that if I want to have two positive solutions, both $x$ should be a positive in their domains, so I got

$$1^\circ \quad 4-2m > 0 \implies \quad x \in \emptyset \quad\text{if}\quad x \geq 2$$ $$2^\circ \quad \frac{4+2m}{3} > 0 \implies \quad x \in (-2, 2) \quad\text{if}\quad x < 2$$

When I take a common part of these two solutions, I finish this task with no $m$ for which this equation has two positive solutions.

However, when I try to solve this graphically, I define $m = \frac12 (x - |4-2x|)$ and sketch the right hand side of this equation

Graph of the function $\quad m = \frac12 (x - |4-2x|)$

$\quad m = \frac12 (x - |4-2x|)$" />

It is clearly visible that two positive solutions are for $m \in (-2, 1)$. Could you please help me in finding the mistake in my consideration? I really want to solve this question without a graphical method.

Answer 1

As a final step you need to solve that :

$$ \begin{cases}4-2m≥2\\ 0<\dfrac {4+2m}{3}<2\end{cases} $$

which yieds $m\in(-2,1)$ .

Answer 2

• If $x=2$ you have $x=2m$, so $m=\frac x2 \in \{ 1\}$
• If $x >2$ you have $x=4-2m$, so $m= \frac{4-x}2 \in (-\infty, 1)$
• If $0<x < 2$ you have $x=\frac {4+2m}{3}$, so $m=\frac{3x-4}2 \in (-2,1)$

You want two positive solutions for a given $m$, and the only non-empty intersection here is between $(-\infty, 1)$ and $(-2,1)$, and is $(-2,1)$.

This shows any $m\in (-2,1)$ will give two positive $x$s, one above $2$ and the other in $(0,2)$, and no other $m$ will do.