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Say we have: $$\sqrt{x+7}=5-x$$

Is it implicitly understood that the following also holds? $$-\sqrt{x+7}=5-x$$

I'm exploring the notion of "extraneous solutions." In this example, solving either equation leads to two results, namely x=2 and x=9.

Standard practice is to check these solutions once they're found by plugging them into the original equation. Thus, x=2 is shown to be the right answer, assuming we're using the first equation above. Meaning, x=9 is extraneous.

But really, x=9 is the solution to the second equation.

What's the right way to think about the square root operator given this discussion? Does the second equation logically follow from the first and, if so, is it right to call the second equation's solution "extraneous?"

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    $\begingroup$ $\sqrt{x+7}$ is always positive (given real inputs and avoiding nonreal outputs). The square root operator $\sqrt{~~~}$ only ever outputs the principle square root, which in the case of positive real inputs will always be a positive real output. $\endgroup$ – JMoravitz Jul 26 '16 at 17:54
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    $\begingroup$ Slight modification to my comment. I must have been half asleep. Replace the word "positive" with "non-negative." It is certainly possible for $\sqrt{x+7}$ to equal zero. $\endgroup$ – JMoravitz Jul 26 '16 at 19:00
  • $\begingroup$ You may find it helpful to read about why $\sqrt{x}$ is a function. $\endgroup$ – N. F. Taussig Jul 26 '16 at 19:16
  • $\begingroup$ For example $z = 1$ has the solution $z = 1$ !!!. Now, squares both sides: $z^{2} = 1$ which has the solutions $z = -1$ and $z = 1$. So, by squaring you introduced an "extraneous solution ( namely, $z = -1$ ). The general procedure is to go back to the original equation $z = 1$ and check all the solutions you got with your "auxiliary equation" $z^{2} = 1$. Obviously, $z = -1$ is not a solution because $-1 \not= 1$. Send it to the trash and keep the 'decent' one $z = 1$. This is the general idea. From now on, 'keep your flags up' whenever you square something. $\endgroup$ – Felix Marin Jul 26 '16 at 20:06
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If what you say is true, then we should be able to add both sides,

$$\sqrt{x+7}-\sqrt{x+7}=5-x+5-x$$

$$0=2(5-x)$$

$$\implies x=5$$

Going back, this is not true in either equation.

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The other answers have their considerable virtues, especially as they do reflect various demographics' sense about this notation and, therefore, about the sense of the question.

Nevertheless, I do think that one cannot rely upon any presumed conventions. First, human conventions cannot change actual mathematical phenomena (in my philosophical world). "Notation has no power over us!" :) But, still, to interpret properly things that other people write, in implicit contexts that are wildly unacknowledged or misunderstood, one must be prepared to accommodate...

The whole "extraneous solutions" notion does indeed have a reasonable sense, but mostly this can be dispelled by dialing-back to ask what the intention is. If the rules are just the formal-game of school-math, ... who knows? It's up to the "teacher" to tell us. That is, one cannot deduce from logical principles the speed limit in a given area, even if one knows that generally that limit is below 35 in residential areas, and so on: maybe it's 20? No way to know without seeing the posts of local legislation, which may have only a "secret" rationality.

I do claim that that last bit of rant is not irrelevant to the question. Namely, in my experience, the school-math drill related to such things asks a misleading question, and (whether by accident or not) the prankish ambiguity or ill-definedness is exactly the intended prank.

That is, if a question's sense, or correctness of answer, depends almost entirely upon human convention, that makes it a wildly different thing from questions about the physical world (which I think includes genuine mathematics). Thus, any narrative that pranks (!) the reader by refusing to explain the context (if only by the place it appears) is sociopathic. Certainly not everything can be explained starting at high-school level, but, still, one can "let on" what "the real deal" is, rather than stone-walling or pretending that "all the important people understand".

So, literally, no, there is no useful, universal convention about signs of square roots, despite the pretensions of many textbooks, which apparently attempt to declare "rules". No. Things are not that easy/clear. In real life, quite often the "query-interaction" is absolutely necessary if one is to know the intent of another... "srsly"...

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Unless explicitely stated we only need to consider the positive square root on the left side. In fact the image of the square root function is the positive reals. Because of all this it's a common practice to plug in the solutions into the equation where we squared both sides.

For example in the formula for solving the quadratic equation we have $x_{1/2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, as only the positive value of the square root is taken.

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Note that $\sqrt { x+7 } =5-x$ and $-\sqrt { x+7 } =5-x$ are different equations.If we multiply both side to $-1$ then we get $$\quad -\sqrt { x+7 } =-\left( 5-x \right) \Rightarrow -\sqrt { x+7 } =x-5\\ \\ $$

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It certainly does not also hold, as it would imply $5x=0$, and $5$ is not a root.

Standard practice is maybe to transform the equation so as to find a list of possible solutions, then plug each of them in the initial equation to check if it's really a solution or a parasit solution.

But better practice is to know a little more on inequalities than the required minimum. For instance, here instead of simply squaring, one may know this logical equivalence $$\sqrt A=B\iff A=B^2 \;\textbf{ and }\; \boldsymbol{ B\ge 0}.$$ Thus one knows at once the root, if any, has to be at most $5$, and eliminates the ‘root’ $9$.

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