Maximising $a$ for a quadratic!

in the above graph $|x-1|=4$ has two solutions.

$|x-1|+a$ means graph is shifted by $a$ units upwards or downwards depending on sign of $a$ , thus to have at least a real solution $a=4$

$\left| x-1 \right| =\pm 4-a$ $\left| x-1 \right| \ge 0$ $\pm 4-a\ge 0$ $a\le \pm 4$ $\max { \left( a \right) } =4$