Confused About Squareroots & Absolute Values $${ x }^{ 2 }=4\\ \sqrt { { x }^{ 2 } } =\sqrt { 4 } \\ |x|=2$$
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?
 A: Let's check this out, backwards, checking to see that the solutions $x = 2$ and $x = -2$ satisfy $|x| = 2$:
\begin{align}
x &= 2: \quad &|2| &= 2 \\ \\
x &= -2: \quad &|-2| &= 2 \\ \\
\end{align}
There are TWO possible "input" values for $x$, such that $|x| = 2$. Note for each of the two inputs, the "output" is the same. Essentially the absolute value function can be described, defined as follows: 
$\quad f:\mathbb R \to \mathbb R,\;f(x) = |x|:$ 


*

*$\;x\mapsto -x\;\;$ if and only if $\;\;x \lt 0$, 

*$\;x \mapsto x,\;\;\;\;$ otherwise


I'll include a graph of the function $$f(x) = y = |x|,\;\;\text{with}\;\;x = -2,\;\; x = 2,\;\; f(x) = y = 2\;\;$$ to help you visualize the symmetric function, its behavior, and why $$\;f(x) = y = |x| = 2\;\implies\;x = -2\;\;\text{or}\;\;x = 2\;:$$
$\qquad\qquad\qquad\qquad$
Graph from WolframAlpha
See also the following posts:


*

*square root of a square is the same as absolute value?

*square roots: positive and negative
A: It's not that the absolute value doesn't give two solutions.
It's just that if you have $±|z|$ this can either be $-(±z)$ or $(±z)$, either way you still only get two cases; namely, $±z$.

If you understand the equation $y=|x|$ then you should be able to see that since $|x|$ takes two values of $x$, both positive and negative, $|-x|=|x|$. So when you put in $x=-2$ you get 2, which is the same as if you plug in $x=2$. So it's symmetric.
A: Absolute value is a single value. But there are two cases, when $x\geq0$, $|x|=x$ and when $x<0$, $|x|=-x$. Since there is $\pm$ sign in your second last line, it does not really matter to separate the two cases. So simply $\pm |z|=\pm z$. 
