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When I was younger I wasn't paying too much attention or the teacher did not make sure we understood how the square root works. Recently I was faced with some problems where having the right knowledge would help a lot. I tried to rethink the stuff, but I got stuck, and the results on internet did not lead to the solutions of my problem

My problem seems to focus on the notations. $\sqrt{c}$ in my mind means any number that squared results "c": $x^2 = c$

As example: $x = \sqrt{4}$ shall have 2 solutions: -2 and 2, because both squared results 4. I understood that $\sqrt{c}$ always means the positive root, so it always have 1 solution. So I take this as a "fake" sqrt, and the 2 is the only solution

In order to comprehend this behavior so I will have faster responses, I told to myself 2 rules:

  • when the square root is generated by myself (like applying square root to both sides of ecuation), use the $\pm\sqrt{c}$ notation
  • when getting rid of it (such as squaring both sides), just get rid of the square root, calculate solutions, and check them as there could be additional invalid ones

So for $x = \sqrt{4}$, I can square both sides and get $x^2 = 4$, where x is $\pm2$, and then check in the initial ecuation and get rid of -2. The problem comes here: I was told that squaring such a "fake" sqrt could result only in true solutions (aka no check required after) by using || to the member inside. It indeed forces the positivity, but it doesn't work with the following example:

$\sqrt{x} = \sqrt{2}$

The only solution there is 2, but using the || knowledge I was talking about: by squaring both sides I get |x| = |2|, then |x| = 2. It nows yeild 2 solutions: $\pm2$, so the promise that it doesn't require any check is vanished. If it requires checks, why is then even required the ||? My method before being reminded about || took care of the problem without any logic overhead

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    $\begingroup$ Additionally to the excellent below answer, $\sqrt{x^2}=|x|$ for all real numbers $x$. This is also something very important. $\endgroup$ Commented May 11, 2020 at 16:57

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For $x\ge0$, we define $\sqrt{x}$ as the unique $y\ge0$ solving $y^2=x$. By definition, then, $\sqrt{x}^2=x$ for all $x\ge0$. There's no need to write $\sqrt{x}^2=|x|$; this is equivalent to the previous equation if $x\ge0$, but if $x<0$ then $|x|\ge0$, while $\sqrt{x}^2$ cannot be said to be $\ge0$. If you're familiar with imaginary numbers, you'll know how $\sqrt{x}$ can be defined for $x<0$ (usually with the requirement that it be $\ge0$ changed to its imaginary part being $\ge0$), and will also know that in that case $\sqrt{x}^2=x<0\implies\sqrt{x}\ne|x|$. And if you're unfamiliar with them or prefer to limit our discussion to real-valued square roots, $\sqrt{x}$ is undefined for $x<0$, so in particular $\sqrt{x}^2$ cannot mean the same thing as $x$.

If $x=y\ge0$, $\sqrt{x}=\sqrt{y}$. If $x^2=y^2$ for $x,\,y\in\Bbb R$, $x=\pm y$ and $|x|=|y|$. And regardless of whether we involve imaginary numbers, if $\sqrt{x}=\sqrt{y}$ then $x=\sqrt{x}^2=\sqrt{y}^2=y$.

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  • $\begingroup$ So if the stuff under $\sqrt{}$ is positive, I never need to use ||? It could be a liniar / quadratic function $\endgroup$ Commented May 11, 2020 at 17:10
  • $\begingroup$ @uIM7AI9S Note the one time I had to include $||$ (or $\pm$) in a conclusion was when we remove a squaring operator, not a square-rooting one: $x^2=y^2\implies|x|=\sqrt{x^2}=\sqrt{y^2}=|y|$. $\endgroup$ Commented May 11, 2020 at 17:13
  • $\begingroup$ This means when I'm upgrading the exponent from the sqrt, I must check solutions, and when I downgrade I must apply || to both sides? $x^2 = 4$ would mean $|x| = |2|$ right? $\endgroup$ Commented May 11, 2020 at 17:24
  • $\begingroup$ @uIM7AI9S Although in that specific example we can just write $|2|$ as its value, $2$. $\endgroup$ Commented May 11, 2020 at 18:01
  • $\begingroup$ Idt the 2 main answers were what I was looking for, but this one did help me deduce myself what I needed $\endgroup$ Commented May 11, 2020 at 18:10
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Your methode is right. But when you want to apply it you have to note before your square the more restrictive condition present in your reasonement :

Example : $\sqrt x = \sqrt 2$ undermeans that $x\geq0$ so you have to memorize that condition in your development. Even if you apply a very complex function to find something else in a given problem you've to take note of and remember the more restrictive condition.

In fact $|x|=2$ will gives you two potential solutions : $2,-2$.

But when remembering your more restrictive condition the only solution is $2$ because $x\geq 0$.

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  • $\begingroup$ So if I have $\sqrt{x^2}$ I can always replace it with only "x" if I do the checks at the end? $\endgroup$ Commented May 11, 2020 at 17:00
  • $\begingroup$ @ulM7Al9S you can only be sure $\sqrt{x^2} = x$ if you know $x \geq 0$ $\endgroup$ Commented May 11, 2020 at 17:02

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