Solve for $x$ in terms of $m$: $|mx+m+2|=x+3$

Question

The problem is about the equation $|mx+m+2|=x+3$. I have to solve this equation but I don't know how to do it. The teacher gave us the solution. Let $S$ be the set of solutions for $x$:

if $m \in (-\infty , -1) \cup (1, +\infty)$, then $S=\{-1, -((m+5)/(m+1))\}$

if $m \in [-1 , 1)$, then $S=\{-1\}$

if $m \in \{1\}$, then $S=[-3,+\infty)$.

I don't even know where to begin nor how to continue. Sorry to bother you all, have a nice evening. Oh, and I'm new over here so sorry if my formatting isn't what it should be, did my best.

Answer 1

First of all, as you observed, it must be $x\geq -3$. In this hypothesis, the given equation is equivalent to that obtained by squaring, that is:

$$ |mx+m+2|^2 = (x+3)^2 $$

or, calculating and simplifying:

$$ (m-1)(m+1)x^2 + 2(m-1)(m+3)x + (m-1)(m+5) = 0. $$

• If $m=1$, equation reduces to $0=0$, which is satisfied for every $x$, for every $x\geq -3$ of course.
• If $m\neq 1$, equation becomes

$$ (m+1)x^2 + 2(m+3)x + m+5 = 0. $$

Two cases:

• If $m=-1$, we obtain $4x+4 = 0$, $\hskip1ex$ i.e. $\hskip1ex$ $x=-1.$
• If $m\neq -1$, we can use the well known formula for quadratic equations, obtaining: $$ x = \frac{-(m+3) \pm \sqrt{(m+3)^2-(m+1)(m+5)}}{m+1} = \frac{-m-3 \pm 2}{m+1} $$

that is $x = -1 \hskip2ex \text{or} \hskip2ex x = \displaystyle -\frac{m+5}{m+1}.\hskip1ex$ Now, we just have to see when the last solution is acceptable, i.e. when

$$ -\frac{m+5}{m+1} \geq -3. $$

This inequality is equivalent to the union of the following two systems:

$$ \left\{ \begin{array}{c} m+5 \leq 3(m+1) \\ m+1 > 0 \end{array} \right. \hskip4ex \text{and} \hskip4ex \left\{ \begin{array}{c} m+5 \geq 3(m+1) \\ m+1 < 0 \end{array} \right. $$

whose solutions are respectively $m\geq 1$ and $m\leq -1$. Remembering that currently is $m\neq 1$ and $m\neq -1$, we obtain that both solutions of the quadratic equation above are acceptable if and only if $m\in [-\infty,-1[\,\cup\,]1,+\infty[$. Otherwise, that is in the interval $]-1,1[$, only $x=-1$ is acceptable.

You can immediately see that the teacher's solution is thus obtained.