Absolute value having $4$ distinct solutions

Question

Let's assume that $||x|+m| = 7$, and that $||x|+m| = 7$ has $4$ distinct solutions. How can we conclude that $m<-7$?

$$||x|+m| = 7\iff |x|+m = \pm 7\iff |x| = \pm 7 - m\iff x = \pm(\pm 7 - m)$$

But I am not sure what we can conclude from here, except $m\neq \pm 7$.

Answer 1

If $||x| + m| = 7$, then $|x| + m = 7$ or $|x| + m = -7$.

If $|x| + m = 7$, then $|x| = 7 - m$, which has two solutions if $7 - m > 0 \implies 7 > m$.

If $|x| + m = -7$, then $|x| = -7 - m$, which has two solutions if $-7 - m > 0 \implies -7 > m$.

Since we want four solutions, we require that $7 > m$ and that $-7 > m$, which implies that $-7 > m$.

Answer 2

If we have four distinct solutions, we require $x\neq 0$ and $|x|+m\neq 0$. From one of your equivalence:

$$ -m\pm 7 = |x| > 0 \phantom{x} \implies \phantom{x} \pm 7 > m $$