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Recently I was watching this interview of Andrew Wiles where a secondary school teacher asked this question:
How do you teach square roots?
He doesn't answer the question so I'd like to ask it here?
Here is another example concerning the widespread confusion about this question; in this clip from University Challenge, he asks "What are the two possible answers to the calculation 5 plus the square root of 4?"
As he says, this is a convention, and you just have to make a choice and stick with it. The usual convention (which makes the square root a function) is to take the positive branch.
The teacher seems a bit confused. He seems to think that there is a contradiction between saying $\sqrt{9} = 3$ on one hand, and on the other hand writing $x^2=9 \implies x=\pm3$ on the other.
If we stick to the convention that the square root takes only positive values, then there are two sensible ways to approach the quadratic equation.
One is to recognize that $\sqrt{x^2} = |x|$, not $x$. So $x^2 =9 \implies |x|=3 \implies x = \pm 3$.
The other way is to factor $x^2=9 \implies x^2-9=0 \implies (x-3)(x+3) = 0 \implies x = \pm3$.
In either case, the $\pm$ does not need to be part of the definition of the square root, and including it makes life more complicated because then square root is not a (single valued) function.
EDIT:
As mentioned in the comments, it is worth making the distinction between square roots and principle square roots. We can define $y$ to be a square root of $x$ if and only if $y^2=x$. Then every positive number $x$ has two square roots, $0$ has only $0$ as its square root, and negative numbers have no real square roots.
The relation $y^2=x$ is not a function of $x$, but if we throw out all the pairs with negative $y$ values we do obtain a function, called the "principle square roots function". This takes a nonnegative real number $x$ as input and returns its nonnegative square root as output.
Another observation: this choice is nice for many reasons, not the least of which is that it is consistent with the definition $x^a = e^{a \log(x)}$ which can be used to define any positive power function (assuming the exponential and logarithm have been defined).
This might help to clarify my remarks above the edit.
Original Answer
I first define that a square root of a number $a$ is a number $x$ such that $x^2 = a$. I then give some examples like: What are the square roots of 4? $2$ and $-2$ because $2^2=4$ and $(-2)^2=4$. I then state that positive real numbers have two square roots, one positive and one negative.
So what does $\sqrt{a}$ mean? There are two choices, right?
Define that the symbol $\sqrt{a}$ always denotes the positive square root of $a$.
Then I given a couple of examples: The square roots of 4 are $\sqrt{4}=2$ and $-\sqrt{4}=-2$. The square roots of 2 are $\sqrt{2}$ and $-\sqrt{2}$.
$n$-th roots are defined in a similar way, as solutions of $x^n=a$. Example: $-2$ is cube root of $-8$ because $(-2)^3=-8$ To entice students, I like to mention that there are actually three cube roots of $-8$: $-2$ and two complex roots (which they will see in their future)
So what does $\sqrt[3]{a}$ mean? There are three choices?
Answer: The symbol $\sqrt[3]{a}$ always denotes the real $3$-rd root of $a$.
Example: $\sqrt[3]{-8} = -2$
The symbol $\sqrt[n]{a}$ denotes the real $n$-th root of $a$, unless there are two real $n$-th roots, in which case pick the positive $n$-th root. Alternatively, break it up according to whether $n$ is odd or even.
Added in response to a question in the comments:
The formulas $(\sqrt{x})^2 = x$ and $\sqrt{x^2} = |x|$ and often confuse students. Here's how I approach it.
Question to students: $(\sqrt{x})^2$ = ?
By definition, $\sqrt{x}$ is the non-negative number such that $(\sqrt{x})^2 = x$.
So $(\sqrt{x})^2 = x$.
Example: $(\sqrt{5})^2 = 5$
Question to students: $\sqrt{x^2}$ = ?
Some students will say $\sqrt{x^2}$ = x$.
Let's test this formula.
Example: $\sqrt{2^2} = \sqrt{4} = 2$. That works.
Example: $\sqrt{(-2)^2}$ = ?. Since $(-2)^2 = 4$, we have $\sqrt{(-2)^2} = \sqrt{4} = 2$. But the suggested formula $\sqrt{x^2} = x$ says that $\sqrt{(-2)^2} = -2$. The formula doesn't work.
Some student will probably now suggest $\sqrt{x^2} = |x|$. If not, you should suggest it.
Now I explain why it works.
By definition, $\sqrt{x^2}$ is the non-negative number such that $(\sqrt{x^2})^2 = x^2$.
So either $\sqrt{x^2} = x$ (because $x^2 = x^2$) or $\sqrt{x^2} = -x$ (because $(-x)^2 = x^2$). How to choose? By definition, pick the non-negative one. If $x \geq 0$, then $\sqrt{x^2} = x$. If $x < 0$, then $\sqrt{x^2} = -x$.
Let's write this as a piecewise function:
$$ \sqrt{x^2} = \left\{ \begin{array}{ll} x & \text{if } x \geq 0\\ -x & \text{if } x < 0\\ \end{array} \right. $$
Does this function seems familiar? This exactly the formula for $|x|$. Therefore $$ \sqrt{x^2} = |x|. $$
NOTE: I am assuming here I have introduced the piecewise formula for the absolute value $|x|$.
The answer in the video was a bit unsatisfying for you, and I understand why. You expect a firm answer.
IRL, when I get this question, I need to quickly determine the level the student is at in her studies. When square roots (or nth roots for that matter) are first discussed, we are only concerned about the positive value. After all, my square with an area of 4 square units has a side that’s a positive number.
But then the students move on the the square root as part of the quadratic equation solving, and we must be mindful of both positive and negative roots.
At some point, the 4th root of 16 has 4 solutions, as the +/-2i is also a root.
It’s for the adult math teacher to understand what level the class is at, and whether that level contains the negative or imaginary root as a solution.
