how many solutions does a square root of a positive number have?

Question

So a friend of mine asked me how many solutions does $\sqrt{49}$, have. I replied with $\pm7$ since I remember when I was younger I didn't get full pot on my test cause I forgot to add the $\pm$ to the answer. But today I got told and learned something, that by definition the square root of a positive integer is always positive. So is it wrong that the answer is $\pm7$, my friend also told me that it's wrong cause I cannot get $-7$ from $\sqrt49$. I tried and I came up with two possible ways (I don't know if these are correct or not)

$\sqrt{49} \longleftrightarrow\sqrt{-7}\sqrt{-7}=\sqrt7\sqrt7\sqrt{-1}\sqrt{-1}=\sqrt{49}i^2=-7$

the other method I used was

$\sqrt49 \longleftrightarrow\sqrt{-7}\sqrt{-7}=(-7)^{1/2}(-7)^{1/2}=(-7)^{1/2+1/2}=(-7)^1=-7$

I know that $x^2=49$ has two possible answers.

Answer 1

There are two numbers that you can square to get 49. You've correctly identified both.

What mathematicians normally mean when they write $\sqrt{x}$ is the positive number that, when squared, gives x. However, this definition breaks down every once in a while and is very context heavy.

I don't think such a distinction is worth argument, as long as you understand what someone else means with their terminology.

Answer 2

Square root of positive real numbers are defined to take the positive root. It is from definition but not from calculation.

$x^2=49$ does have $2$ roots, but we will only take the positive one for square root.

It is done, to preserve the property that $sqrt$ as a function, instead of a multi-valued mapping.

Answer 3

$\sqrt{x}$ for real $x$ is defined to be the positive real solution of the equation $z^2 - x = 0$.

However, the question your friend asked is ill-formed, because asking "how many solutions $\sqrt{47}$ has" is not a proper question. There is no equation to solve, thus there are no solutions to speak of.

Answer 4

Back in the day when I was teaching in high school I remember a student who had just arrived from a foreign country (can't remember which country) telling me that where he was from, they used to write things like $$\sqrt{4}=\{-2, 2\}$$

which does make sense, in a way. So you are not plainly wrong. But this is not how you define a function in the classic, western sense. A function can take only one value at a time. If there are several choices that would seem natural, you have to choose one of them, once and for all, if you are to define a function.

Note however that there is a western notation that will output things like $\{-2, 2\}$. If you let $f(x)=x^2$ denote the square map, then we use the notation $f^{-1}(49)$ to denote the set of all inputs that will output $49$ via the function $f$. Therefore $$f^{-1}(49)=\{-7, 7\}$$

Answer 5

1) $\sqrt {49}$ is not a question and does not have any solutions.

2)"How many square roots does 49 have?" is a question and the answer is "2; 7 and -7".

3) "What are the square roots of 49" is another question and the answer to that is "7 and -7".

4) And finally a fourth question is "What is the positive square root of 49" and the answer to that is "7".

Notice all four of those are different (non)questions! And different questions have different answers.

So the answer to the question:

"What is $\sqrt 49$" is "By definition the positive square root of 49" is "$7$ or $-7$.

And so the one solution to $x = \sqrt{49}$ is just like the solution to $x = 3+4$. It's $x = 7$. That's it.

But the answer to the different question:

"What numbers when squared give you $49$" is "The square roots of $49$" is "There are two of them $7$ and $-7$".

ANd the solution to $x^2 = 49$ is:

$x^2 = 49 \iff x^2 - 49 = 0 \iff (x-7)(x+7) = 0 \iff x -7 =0$ OR $x+7 = 0 \iff x = 7$ OR $x = -7$. So $x = \pm 7$.

Or to do it more quickly:

$x^2 = 49$

$x = \pm\sqrt {49}$

$x = \pm 7$.

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The upshot is:

1) Every positive real number has two square roots; a positive value and a negative value, and the absolute value of these two roots are equal (i.e. the are equal in magnitude but of opposite parity.)

2) The notation $\sqrt {foo}$ is by definition the positive square root. And so the notation $-\sqrt{foo}$ is by construction the negative square root.

So for $k > 0$ the two solutions to $x^2 = k$ are $x =\pm \sqrt k$ where $-\sqrt{k} < 0 < \sqrt{k}$.