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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.

  • 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. – fleablood Apr 20 '17 at 00:35
  • @fleablood I think in my original question it was $+$ sign. –  Apr 20 '17 at 00:37
  • Oh, in that case, no, you can't say it $\sqrt{100} + 10 = 20 \ne 0$ so it is not a solution. That's all there is to it. – fleablood Apr 20 '17 at 00:38
  • No, yo can't say that – The Dead Legend Apr 20 '17 at 00:40
  • @fleablood But isn't $(-10)^2 = 100$ ? I get that $10$ is the square root of $100$ by convention, that is because to fit $\sqrt{x}$ in the definition of function but here will it matter if I don't consider $\sqrt{x}$ as a function ? –  Apr 20 '17 at 00:40
  • Squaring both sides adds extraneous solutions. $x = 5$ has only one solution. Squaring both sides $x^2 = 25$ add the extraneous, and wrong, "solution" $x = -5$ which is not a solution. – fleablood Apr 20 '17 at 00:40
  • If the range of $\sqrt x$ is strictly positive, how can $\sqrt x$ equal a negative number? – Simply Beautiful Art Apr 20 '17 at 00:41
  • @SimplyBeautifulArt The range is positive (not strictly) if it is function. But here in this question it does not matter if I don't consider it to be function right ? –  Apr 20 '17 at 00:42
  • Of course, but that is totally irrelevant. $x = 2 \implies x^2 = 4$. $x^2 = 4$ has two solutions. Squaring both sides does not create an equivalent equation. It makes a weaker solution. The solutions to $x = k$ are solutions to $x^2 = k^2$ but the sollutions to $x^2 = k^2$ are not solutions to $x = k$. – fleablood Apr 20 '17 at 00:44
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    Please see: https://math.stackexchange.com/questions/809424/can-the-square-root-of-a-real-number-be-negative – HEKTO Apr 20 '17 at 00:46
  • I love to know the reason of the downvote. –  Apr 20 '17 at 00:47
  • @fleablood I do know that but that is irrelevant here because 100 is a solution to $\sqrt{x} + 10 = 0$ if we consider its negative root. –  Apr 20 '17 at 00:50
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    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. – JMoravitz Apr 20 '17 at 00:50
  • @JMoravitz I think the question was not phrased properly. I get that we dont take negative root but that is because of definition of functions. What should I use if I want to consider both roots ? –  Apr 20 '17 at 00:54
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    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. – fleablood Apr 20 '17 at 00:56
  • @fleablood Ok then do I have to use $\pm \sqrt{x}$ everywhere if I have to use negative roots as well ? –  Apr 20 '17 at 00:58
  • Then you want to solve $\pm \sqrt{x} + 10 = 0$. So $\pm \sqrt{x} = -10$. $\sqrt{x} =-10$ is simply impossible. So $-\sqrt{x} = -10$ and $ \sqrt{x} = 10$ and $x = 100$. So $x = 100$ is a (only) solution to $\pm \sqrt{x} + 10 = 0$. – fleablood Apr 20 '17 at 01:00
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    Yes you do. <><> – fleablood Apr 20 '17 at 01:01
  • @fleablood Well I think something positive come out of this question. I did not know $\sqrt $ stands for +ve roots. I think I can sleep well tonight. –  Apr 20 '17 at 01:03

4 Answers4

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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.

Ángel Mario Gallegos
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fleablood
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    I think to be precise you need to mention that $\sqrt{x}$ is called principal square root, which is non-negative by definition. Please see: http://mathworld.wolfram.com/PrincipalSquareRoot.html – HEKTO Apr 20 '17 at 03:19
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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.

Alex Li
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$x=100$ 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.

ASKASK
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$√(x)+10=0$

$√(x)=-10$

$√(x)^2=(-10)^2$

$x=100$

$√(100)+10≠0$

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

uncertainty
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