Does $2^2=4$ imply $2=\pm \sqrt{4}$?

Question

I read the square root property from the book, College Algebra by Raymond A Barnett and Micheal R Ziegler that,

The square root property says,

If $A^2=C$ then $A=\pm \sqrt{C}$

I took the equality,

$2^2=4$

$\implies 2=\pm \sqrt{4}$

$\implies 2=\pm {2}$

but, How $2$ can be equal to $-2$?

Answer 1

Note that $\sqrt4=\pm2$ means that $\sqrt4 = +2 \;\lor -2$, not$\sqrt4 = +2 \land\; -2$. You must have misread the $\pm$ notation, which reads as "plus OR minus", not "plus and minus". However, $2 = \pm2$ is certainly true. Note that $\lor$ is the logical OR.

Answer 2

$2$ must be equal to $2$ OR to $-2$. Just one of them, not both.

Answer 3

Let's see how "If $A^2=C$ then $A=\pm \sqrt{C}$" is obtained:

$A^2=C \Rightarrow A^2-C=0 \Rightarrow (A-\sqrt{C})(A+\sqrt{C})=0 \Rightarrow$ $A-\sqrt{C}=0$ or $A+\sqrt{C}=0 \Rightarrow A_1=\sqrt{C}$ or $A_2=-\sqrt{C} \Rightarrow A=\pm \sqrt{C}$.

Now let's see what happens to "$2^2=4$":

$2^2=4 \Rightarrow 2^2-4=0 \Rightarrow (2-\sqrt{4})(2+\sqrt{4})=0 \Rightarrow$ $2-\sqrt{4}=0$ or $2+\sqrt{4}=0 \Rightarrow 2=\sqrt{4}$ or $2\ne-\sqrt{4} \Rightarrow 2=\sqrt{4}$.