Confused About Squareroots & Absolute Values

[InvalidID]

$${ x }^{ 2 }=4\\ \sqrt { { x }^{ 2 } } =\sqrt { 4 } \\ |x|=2$$
According to my professor, in the above case, the absolute value gives two solutions: ##x=±2##
Consider the discriminant in the quadratic formula: $$x=\frac { -b±\sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ Let\quad { z }^{ 2 }={ b }^{ 2 }-4ac\\ ±\sqrt { { b }^{ 2 }-4ac } \\ =±\sqrt { { z }^{ 2 } } \\ =±|z|\\ =±z$$
However, according to my professor, in this case, the absolute value gives only one solution: ##|z|=z##

How come the absolute value sometimes gives one solution and sometimes it gives two solutions?

 

[Mark44]

$${ x }^{ 2 }=4\\ \sqrt { { x }^{ 2 } } =\sqrt { 4 } \\ |x|=2$$
According to my professor, in the above case, the absolute value gives two solutions: ##x=±2##
Consider the discriminant in the quadratic formula: $$x=\frac { -b±\sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ Let\quad { z }^{ 2 }={ b }^{ 2 }-4ac\\ ±\sqrt { { b }^{ 2 }-4ac } \\ =±\sqrt { { z }^{ 2 } } \\ =±|z|\\ =±z$$
However, according to my professor, in this case, the absolute value gives only one solution: ##|z|=z##

This makes no sense to me. If z is ≥ 0, then |z| = z. OTOH, if z < 0, then |z| = -z.

How come the absolute value sometimes gives one solution and sometimes it gives two solutions?