TITLE: Is $\sqrt{64}$ considered $8$? or is it $8,-8$?
QUESTION [56 upvotes]: Last year in Pre-Algebra we learned about square roots. I was taught then that
$\sqrt{64}=8$ and $\sqrt{100}=10$, which I understood and accepted. I was also taught that $\pm\sqrt{64} = 8,-8$ because both of those numbers squared is 64, which I also get.
But this year, with a new school and teacher in a different state, our teacher is telling us that:
$\sqrt{64}=8,-8$ and $\pm\sqrt{64}$ also is $8,-8$. The way to get the positive root of something is:
$+\sqrt{64}=8$
And these seem to contradict each other. I was always taught that a regular square root returned a positive number and only a positive number, but now my teacher is saying a regular square root gives two numbers, and considering the square root of a number $n$ is defined as
$y^2=n$ I see where he is coming from.
Upon researching this Wikipedia says:

For example, $4$ and $−4$ are square roots of $16$ because $4^{2} = (−4)^{2} = 16$

And Wolfram MathWorld says:

Note that any positive real number has two square roots, one positive and one negative. For example, the square roots of $9$ are $-3$ and $+3$

But on the other side, Wolfram Alpha, when given "The square root of 9" gives only 3.

So, which is right? Is $\sqrt{64}$ considered $8$? or is it $8,-8$?

REPLY [2 votes]: Ultimately, it's a matter of notation. There is no right or wrong answer.

Over time, for better or for worse, mathematicians have adopted the radical sign to mean the principle (nonnegative) square root. In reality, true square roots are either positive or negative.
The first thing to understand is that when you are discussing true square roots, $ \sqrt{25} \ne 5 $. In mathematics, having equality means having one and only one distinct solution. The true $ \sqrt{25} $ has two distinct solutions, so there is nothing equal to it.
Consider the following,
$$ x^2 = 100 \Longleftrightarrow x = \sqrt{100} \Longleftrightarrow x = 10 \lor x = -10 $$
The $ x = 10 \lor x = -10 $ at the end is key here. Both $ x = 10 $ as well as $ x = -10 $ satisfy the first equation, so the assertion is true. However, it is untrue to say (in any context) that $ x = 10 \land x = -10$, since it implies that $ 10 = -10 $, which is obviously false.
Simultaneously, $ x = 10 \lor x = -10 $, but $ x \ne 10 $ (and $ x \ne -10 $). That's not a mistake. When combined, either $ x = 10 $ or $ x = -10 $ are true, but neither are true alone. Both statements directly contradict each other, therefore neither contract each other.
In the equation $ \sqrt{25} = \pm{5} $, plus-or-minus is being used to represent a value with two solutions, i.e. $ \pm{5} = 5 \lor \pm{5} = -5 $. $ \pm{5} $ is not a number, but rather, a value with two solutions; $ \pm{5} \in \{-5, 5\} $, nothing more.
Using plus-or-minus on a true square root does nothing, since the square root is already positive or negative; $ \pm{(\pm{n})} \in \{-(\pm{n}), (\pm{n})\} = \{\pm{n}, \pm{n}\} = \{\pm{n}\} \therefore \pm{(\pm{n})} = \pm{n} $. By the same token, negating a square root or making it positive have no effect.

In the quadratic formula,
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, ax^2 + bx + c = 0 $$
The $ \pm $ can be substituted with either $ + $ or $ - $ and it still holds true. The quadratic formula works regardless of whether you believe $ \sqrt{} $ means principal or true square root. Neat, huh?