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When square rooting something, do we have to put the $\pm$ sign as soon as the $\sqrt{}$ sign appears? For example, consider the following equation:-

$$x+1=5$$

Process 1:

$$\implies \sqrt{x+1}=\pm\sqrt{5}$$

Process 2:

$$\implies \sqrt{x+1}=\sqrt{5}$$

Which of the following is correct?

lone student
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tryingtobeastoic
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    The correct one would be #3: $;x+1=5 \iff x+1 \ge 0 ;\land; \sqrt{x+1} = \sqrt{5},$. – dxiv Sep 02 '21 at 04:53
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    @dxiv But it's useless and #2 is correct. If $x+1=5$, we know $x+1\ge0$, and we can conclude $\sqrt{x+1}=\sqrt{5}$. Note that it's a $\implies$ in the question. – Jean-Claude Arbaut Sep 02 '21 at 04:59
  • @Jean-ClaudeArbaut I don't see why a full equivalency would be less useful than a one-way implication. On the contrary, I find it safer so that one doesn't lose sight of the original problem to solve. – dxiv Sep 02 '21 at 05:09
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    @dxiv Ok, maybe not useless. I should have written not mandatory. The OP is using implications, and you write The correct one, as if the other two were wrong, while one of them is absolutely correct. – Jean-Claude Arbaut Sep 02 '21 at 05:11
  • @Jean-ClaudeArbaut I should have called the equivalency "the better one" rather than "the correct one" to avoid confusion. Implication #2 is correct in and by itself, but one has to remember it's one way, otherwise it's easy to fall for $\require{cancel},a=b^2, a+|b|=\sqrt{a}-1 \implies a+\bcancel{\sqrt{a}}=\bcancel{\sqrt{a}}-1\implies a=-1,$. – dxiv Sep 02 '21 at 05:23
  • @dxiv Yes, when solving an equation by implications, one always has to check the solutions are really solution to the original equation, and were not introduced afterwards. It's thus better to solve by equivalences. You should make this another answer, as it's indeed easy to fail on this. – Jean-Claude Arbaut Sep 02 '21 at 05:26

3 Answers3

3

Only the second one is correct.

If $a=b$, then $\sqrt{a}=\sqrt{b}$. Note that $x\to\sqrt{x}$ is a function from $[0,+\infty)$ to $[0,+\infty)$: you can only take the square root of a nonnegative number, and the result is always a nonnegative number.

Your process 1 is referring to something else. You write $a=\pm b$ as a shortcut for $a=b$ or $a=-b$, and both are supposed to be valid [1]. And it doesn't really appear because of the square root.

For example, say, $x^2=4$. Then, taking square root,

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

$$|x|=2$$

And only then,

$x=2$ or $x=-2$, that is $x=\pm2$.

You could also write:

$$x^2-4=(x-2)(x+2)=0$$

Hence $x=2$ or $x=-2$, that is $x=\pm2$.

You have to remember that $\sqrt{x^2}$ is not $x$, but $|x|$.

[1] One might want to say that because "$2=2$ or $2=-2$" is true, one could write $2=\pm2$, but I don't think it's ever used that way.

Jean-Claude Arbaut
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  • "you can only take the square root of a non-negative number, and the result is always a nonnegative number."

    I agree with your first sentence, but I don't agree with your second sentence. I know that we can only take the square root of a non-negative number. However, the result is not always non-negative. For example, $\sqrt{4}=\pm 2$.

     – tryingtobeastoic Sep 02 '21 at 05:13
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    @AbuSafwanMdFarhan A positive number has two different square roots, but the square root function is always the positive one. That is, the equation $x^2=4$ has two solutions, but $\sqrt{4}$ is always $+2$, and the two solutions of $x^2=4$ are really $\sqrt{4}$ and $-\sqrt{4}$. – Jean-Claude Arbaut Sep 02 '21 at 05:15
  • @Jean-ClaudeArbaut Why is it important to talk about the square-root function then? For finding the correct solution to a quadratic equation, we must pay attention to both the positive and negative square roots. – tryingtobeastoic Sep 02 '21 at 05:19
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    @AbuSafwanMdFarhan Because I need the square root function to write one of the square roots (i.e. the positive one). When I write $x^2=2$ hence $x=\sqrt2$ or $x=-\sqrt2$, I know exactly what $\sqrt2$ is referring to, and it's a positive number. – Jean-Claude Arbaut Sep 02 '21 at 05:21
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    @Abu: We should and we do. As of the square root function, it is the function $\sqrt{x}\equiv x^{\frac12}$ (defined in the principal branch $\mathbb{R}^+$) and it is useful in several real life situations as well as maths. – ultralegend5385 Sep 02 '21 at 05:24
  • @Jean-ClaudeArbaut Gotcha! So, my process 1 isn't completely incorrect though, right? Because you say this yourself," And only then,

    $x=2$ or $x=−2$, that is $x=±2$"

     – tryingtobeastoic Sep 02 '21 at 05:26
  • @AbuSafwanMdFarhan No, $\sqrt{x+1}$ and $\sqrt{5}$ are both positive and $\sqrt{x+1}=-\sqrt{5}$ can't be correct. You should read the last sentence of my answer. – Jean-Claude Arbaut Sep 02 '21 at 05:28
  • @AbuSafwanMdFarhan Another reason to prefer a function: a function has one, and only one value. Dealing with multivalued functions is a mess. – Jean-Claude Arbaut Sep 02 '21 at 05:31
1

Short answer.

Use the definition of the square root:

$$\sqrt{x^2}=|x|≥0$$

and

$$x^2=y^2\iff |x|=|y|.$$

Note that, $x+1=5\implies x+1≥0$.

Then you have

$$\begin{align}&x+1=5,\thinspace x+1≥0\\ \iff &\left(\sqrt{x+1}\right)^2=\left(\sqrt 5\right)^2\\ \iff &|\sqrt{x+1}|=|\sqrt 5|\\ \iff &\sqrt{x+1}=\sqrt 5.\end{align}$$

lone student
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0

The $\pm 3$ symbol, for example, literally means $+3$ OR $-3$.

It only works when we work with something like $x^2 = 5$ then taking square root of both sides we got $\pm x = \sqrt 5$

Note that originally the $\pm$ sign is on the left, not on the right, since $$\sqrt{x^2} = \pm x.$$ Then the "hidden step" is multiplying by negative and get $x = \pm \sqrt 5$. Note also that here we deal with two equalities instead of one. It just looks like only one.

But $\sqrt{4} = 2$ not $\pm 2$ since $\sqrt{x}$ mean "the non-negative square roots of $x$"

Azlif
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    Note that for $x\in \mathbb{R},\sqrt{x^2}=|x|$. So $$\sqrt{x^2}=\begin{cases}x, \mbox{ if }x\ge 0\-x, \mbox{ if }x<0\end{cases}$$ – Ixion Sep 02 '21 at 05:04