Is it wrong to say that 100 is solution of $\sqrt x +10=0$?

I know that range of $\sqrt{x}$ is $[0, \infty)$ by convention. This convention is because of definition of a function. So if I consider $\sqrt x+10=0$ as a simple equation then can I say hundred is a solution to the equation?

I think I can say that because if I square both sides I get $x=100$ and also if I consider $\sqrt x+10=0$ just as a equation then there is no need of thinking about functions and the convention of omitting negative roots.

  • $\begingroup$ You can say it because $\sqrt{100} - 10 = 0$. So 100 is a solution. That's all there is to it. Why wouldn't you be able to say that. $\endgroup$ – fleablood Apr 20 '17 at 0:35
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    $\begingroup$ Please see: math.stackexchange.com/questions/809424/… $\endgroup$ – HEKTO Apr 20 '17 at 0:46
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    $\begingroup$ It wasn't me who downvoted, but I don't disagree with the downvote. Some of the suggested reasons for downvoting: "The question does not show any research effort; it is not clear or not useful". The shortest amount of research into the question of your own will tell you that $\sqrt{x}$ is always positive or zero, and that a positive number plus another positive number is always positive, implying that $\sqrt{100}+10$ is certainly not zero. We only consider the positive root in any context that the $\sqrt{~}$ symbol is used, regardless of whether we are calling it a function or not. $\endgroup$ – JMoravitz Apr 20 '17 at 0:50
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    $\begingroup$ Ah.. But $\sqrt{n}$ is NOT the negative square root. It is, by DEFINITION, the *positive square root. $\sqrt{n} = -10$ does NOT mean $(-10)^2 = n$. It means $(-10)^2 = $ AND $-10 \ge 0$. Which is simply not true. $\endgroup$ – fleablood Apr 20 '17 at 0:56
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    $\begingroup$ Yes you do. <><> $\endgroup$ – fleablood Apr 20 '17 at 1:01

The solutions to $f(x) = k$ are a SUBSET of the solutions to $f(x)^2 = k^2$ but not all the solution to $f(x)^2 = k^2$ are solutions to $f(x) = k$. Squaring both sides of an equation add extraneous solutions.

$\sqrt{x} + 10 = 0$

$\sqrt{x} = -10$ Doesn't just mean that $(\sqrt{x}^2 = (-10)^2$. It ALSO means that $\sqrt{x} = -10 < 0$.

When we square both sides we LOSE information.

$\sqrt{x}^2 = (-10)^2$

$x = 100$ but we have completely LOST that $\sqrt{x} < 0$.

Any $\sqrt{100} + 10 = 10 + 10 = 20 \ne 0$. SO it simply DOESN'T work.

Consider this:

$x =2 $ has one solution. Square both sides and you get $x^2 = 4$. Which has TWO solutions!!! Where did that solution $x = -2$ come from?

It came because when we squared both sides we added invalid extraneous solutions.

Solutions to $x= 2$ is $\{2\}$. Solutions to $x^2 = 4$ has solutions $\{2,-2\}$ and $\{2\} \subset \{2,-2\}$. But it doesn't go the other way. It only goes one way.


Consider the equation x^2 = 100

When we take the square root of the equation to solve it, we would write

x = ± sqrt(100)

Notice that there is a ± sign in front of the square root. The reason that is there is precisely because sqrt(x) is defined as the principle square root.

So, when you evaluate sqrt(100) + 10, you get 10 + 10 = 20, proving your equation false.


$x=100$ is most definitely a solution to $\sqrt{x}-10=0$, because if you plug $x=100$ into the equation, you get $\sqrt{100}-10=0$, or $10-10=0$, which is true.

  • $\begingroup$ I think the OP changed the + sign to the minus sign, which he must have meant all along. $\endgroup$ – Mark Fischler Apr 20 '17 at 0:35
  • $\begingroup$ @MarkFischler edited accordingly $\endgroup$ – ASKASK Apr 20 '17 at 0:35
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    $\begingroup$ @MarkFischler No that was not my edit. $\endgroup$ – A---B Apr 20 '17 at 0:36






Thus, since $√(100)+10$ does not equal zero, this equation has no solution.


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