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We have the equation $$\frac{\sqrt {x}}{2}=-1$$ I proceed as follows $$\sqrt {x}=-2 $$ $$x=4$$ Which does not certainly solve the equation. Where did I go wrong?

Blue
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    Whenever you square both sides of an equation, you risk getting solutions to the new equation that don't satisfy the original. That's just how squaring works...It's an if-then kind of operation, not an only if kind. – aman Mar 13 '21 at 07:04
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    This won't have any solution because there aren't any real numbers that can satisfy this equation. – JonDoe Mar 13 '21 at 07:08
  • Even in the set of complex numbers $\Bbb C$, the square root (on the principal branch) is never a negative real: it is either a non-negative real or an element of $\Bbb C\setminus\Bbb R$, so there isn't any solution in $\Bbb C$ as well. – Prasun Biswas Mar 13 '21 at 07:10
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    @Justin Where is my reasoning flawed? –  Mar 13 '21 at 07:16
  • I think you already solved the equation, because square root of any number $n$, has two solutions just as the quadratic $x^2-n=0$, now for which field does the $\sqrt{4}$ represents $-2$ alone – Aderinsola Joshua Mar 13 '21 at 07:19
  • @Feynstein: Your reasoning isn't flawed. It shows that if there is a solution $x$ to the original equation, it must also satisfy the squared equation giving $x=4$ but $\sqrt 4=2$, hence no solution exists for the original equation, even in $\Bbb C$ – Prasun Biswas Mar 13 '21 at 07:20
  • @PrasunBiswas Assuming mathematics is a self-consistent field, how can a solution solve one equation, but not the same when represented differently? –  Mar 13 '21 at 07:22
  • @Feynstein: they are not the same equation: the squared equation introduces extraneous solutions because the map $x\mapsto x^2$ is not injective: for example, consider $x=2$ but its squared equation $x^2=4$ has an extraneous solution $x=-2$ – Prasun Biswas Mar 13 '21 at 07:25
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    @PrasunBiswas So I cannot square both sides, while keeping the equation invariant? –  Mar 13 '21 at 07:28
  • @Feynstein: only the forward implication is true, ie, $f(x)=g(x)\implies f^2(x)=g^2(x)$, not the other way around. – Prasun Biswas Mar 13 '21 at 07:31
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    @PrasunBiswas They never teach us this in school. Anyways, thanks. –  Mar 13 '21 at 07:33
  • Related: Proving square root of a square is the same as absolute value, Why is $\sqrt{x^2}= |x|$ rather than $\pm x$?. – Martin R Mar 14 '21 at 06:17
  • @lonestudent Yeah, I thought in that way. Hence the question. –  Mar 14 '21 at 06:20
  • Please provide a title that more-accurately conveys the content of your question. Informative titles give you the best chance of attracting answerers; vague and/or generic titles frustrate users searching the site for help on a specific topic. Thanks! – Blue Mar 14 '21 at 06:26
  • @Feynstein: If you know that already then why do you write “And indeed we see that $\sqrt 4 =\pm 2$” ? – It is still unclear to me what the actual question is. Are you asking why a function cannot take two values for a given argument? – Martin R Mar 14 '21 at 06:28
  • @MartinR Yes. Absolutely! –  Mar 14 '21 at 06:29
  • @Feynstein: You may want to look up the definition of a function: “Intuitively, a function is a process that associates each element of a set X, to a single element of a set Y.” – Martin R Mar 14 '21 at 06:31
  • @Martin So 4 is a solution if we agree to accept the negative square root as the output for square root function? –  Mar 14 '21 at 06:33
  • It would have been preferable to edit your previous question in order to clarify why the explanations so far were unsatisfactory to you. I think it is possible for you to flag your own question to ask a moderator to merge the two questions together. – David K Mar 14 '21 at 06:33
  • @DavidK Ok flagging to merge. Everybody flag to merge! Also, my previous post got a single answer, after which it was closed. The answer was not satisfactory. –  Mar 14 '21 at 06:35

1 Answers1

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By definition of the square root :

  • If $\sqrt x=a$, then

$$\begin{cases} x=a^2 \\ a≥0 \end {cases}$$

Now, you can check your solution.

Note that, $-2$ is the real-valued root of $4$. But, we must take only principal square root, which equals to $2.$

Small supplement:

  • If $\sqrt x=a$, then you get $x=a^2$. This is correct. But, if the condition $a≥0$ doesn't hold, then the solution doesn't exist.
lone student
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