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I'm little puzzled with square roots basics which says that square root of a squared number is absolute value of that number. I was in a impression that it would have both positive and negative roots.For example

The Basic says: $\sqrt{2^2} = |2| = 2$

While I think that: $\sqrt{2^2} = \sqrt{4} = 2$ or $-2$.

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  • $\begingroup$ What have you tried to resolve the confusion? $\endgroup$
    – BadAtAlgebra
    Apr 7, 2019 at 21:50
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    $\begingroup$ @BadAtAlgebra Offer conceptual revelation or log off and quit wasting time as a non-contributor critic on a math Q&A site. Why are you here? $\endgroup$
    – CogitoErgoCogitoSum
    Nov 25, 2020 at 23:34
  • $\begingroup$ In order to be a function it has to return one and only one value. The principle square root is the one we chose to use in defining the function. It is true that both 2 and -2, when squared, make 4. So $x^2 = 4$ is solvable with $x=\pm \sqrt{4} = \pm 2$. But that is not saying that $\pm 2$ is the result of $\sqrt{4}$, it is saying that $\pm 2$ is the result of $\pm\sqrt{4}$. The $\pm$ is already there. $\endgroup$
    – CogitoErgoCogitoSum
    Nov 25, 2020 at 23:37
  • $\begingroup$ Sequentially, squaring a value would produce a positive result. And the subsequent root determination should account for all the possible values. This is problematic programmatically since the second operation only knows about the squared root. A traditional, recursive parsing algorithm could not successfully interpret your expression. Math has its exceptions, and this must be one of them. $\endgroup$
    – bvj
    Sep 30, 2021 at 6:14

4 Answers 4

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$\sqrt{a^2}=|a|=a$. This is because the square root function returns the principal (positive) root.

Note that both $a\times a$ and $-a\times-a$ equal $a^2$ though.

You may be wondering when do we use the above fact. An example is when solving a quadratic equation:

Solve for $a:$

$(a^2-2)=0\implies a^2=2\implies a=\sqrt{2}$ or $a= -\sqrt{2}$

So to summarize, when dealing with the square root, the principal root is returned, while solving for variables, you can consider all values for which the equation holds.

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  • $\begingroup$ The OP made no reference to functions. And your justification the root must be positive because the function returns only a positive result is meaningless. $\endgroup$
    – bvj
    Sep 30, 2021 at 6:23
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The square root function is defined such that it returns the positive root of a square, otherwise it wouldn't be a function.

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  • $\begingroup$ I'm just wondering, how we define a square. even a number like 4 whose roots can be both 2 and -2, can also be written in the form of a square. $\endgroup$
    – vikram
    Nov 13, 2013 at 2:44
  • $\begingroup$ The square of $a$ is defined as $a\times a$. So $\sqrt{(-2)^2}=\sqrt{4} = 2$. This indeed means that the square root function is not the inverse of the square function, when the domain is negative. $\endgroup$
    – user85798
    Nov 13, 2013 at 14:42
  • $\begingroup$ You also introduced the function limitation that the OP didn't mention. $\endgroup$
    – bvj
    Sep 30, 2021 at 6:24
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Compare the graph of $y^2 = x$:

y^2=x

with that of $y=\sqrt x$:

y=sqrt(x)

(Pictures courtesy of Wolfram Alpha.)

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  • $\begingroup$ Meaningless function projections. The OP made no reference to functions. $\endgroup$
    – bvj
    Sep 30, 2021 at 6:28
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It is a function for the negative choice too. So the idea that the positive root is chosen to form a function is arbitrary. The better approach is to respect, in any root problems, the possibility of multiple answers. You can then apply your own preference to fit your application. But that is your choice, not the math's. Its like quantum mechanics, where more than one simultaneous state is possible, but the observation reduces to a specific instance.

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  • $\begingroup$ The OP made no reference to functions. FWIW, I did not down-vote. $\endgroup$
    – bvj
    Sep 30, 2021 at 6:31

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