2

$\sqrt{a}=\pm a$ for any $a$. $x^2$ always removes the negative, meaning that it will result in a positive number for $a$, but that doesn’t change the ambiguity of the square root operation. Thus I would think that $\sqrt{x^2}=\pm x$, where you select which output to use based on the problem in question, or leave it ambiguous if there is not enough information.

Yet in usage, it seems like the positive root is assumed if $a$ can be written as the square of something (ie $y=\sqrt{x^2+4x+4}=\sqrt{(x+2)^2}$ is assumed to have range $y\ge 0$). This would seem to contradict the definition, since ANY $a$, be it an equation or a number, can be considered the square of something and thus any use of the square root could be considered to be acting on a square.

What is going on here?

Elliot
  • 211
  • 2
    $\sqrt{x}$ is a function, so it cannot take two values. We've chosen $\sqrt{x}$ to mean the positive real root, for the same reason we chose positive angles to sweep counterclockwise. – Randall Jun 01 '18 at 03:25
  • 2
    Related: https://math.stackexchange.com/questions/26363/square-roots-positive-and-negative, https://math.stackexchange.com/questions/1448885/square-root-confusion (and surely many other questions on this site). – Hans Lundmark Jun 01 '18 at 03:29
  • The thing to note is, what satisfies an equation, ie $x^2 = 1 \implies x = \pm 1$ is not the same as the output of a function, namely the square root, which is restricted to positive real numbers to keep it a function. – Andrew Li Jun 01 '18 at 03:30
  • $\sqrt{}$ is a mark on a piece of paper. Nothing more. Nothing less. People use it to convey a concept that they universally agree upon. The concept that people agree upon is that if $x$ is a non-negative real number we use $\sqrt{x}$ is the positive square root of $x$. So the answer to why $\sqrt{x^2}= |x|$ and not $\pm x$ is because we say it is. That is all there is to it. – fleablood Jun 01 '18 at 05:00
  • 1
    Thing to keep in mind. $\pm k$ is not a number or a value it is notation to mean what we are referring could be one of two possible values but we do not know which. So if $w$ is a number so that $w^2 = x^2$ then $w = \pm x$ because we know $w$ is one or the other of $x$ or $-x$ but we don't know which one. But when we write $\sqrt{x^2}$ we do not mean "some number that when squared is $x^2$ but we don't know which one". We mean "the unique non-negative number that when square is $x^2$ and we do know which one; the non-negative one". Why does it mean that? Because we say it does. – fleablood Jun 01 '18 at 05:13

1 Answers1

4

If $x\in\mathbb{R};x\ge 0$, then $\sqrt{x}$ is defined as "the non-negative real number which, when squared, equals to $x$.

Saying this is correct:

$\sqrt{x^2}=|x|=\begin{cases}x, \text{if }x\ge 0\\-x, \text{if }x\le 0\end{cases}$

I will give you another example first before considering your question:

$x^2-1=0\Leftrightarrow x^2=1\Leftrightarrow x=\pm 1$

For this case, $x=1$ and $x=-1$ are both true for the equation.

However:

$\sqrt{x^2}=|x|=\pm x$

$\sqrt{x^2}=|x|$ is true, however $|x|=\pm x$ is not true, because $|x|=x$ is only true if $x\ge 0$ and $|x|=-x$ is only true if $x\le 0$, they cannot both be correct at the same time (except for $x=0$, but saying $|0|=\pm{0}=0$ is quite pointless).

user061703
  • 3,877
  • $x^2 = 1 \rightarrow \sqrt{x^2} = x = \sqrt{1} = \pm 1 \neq |x|$. You then state the opposite in the following example by stating it as a definition, but in contradiction to the prior definition. This is precisely the point I am making: The definition seems to change arbitrarily. – Elliot Jun 01 '18 at 14:04