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If the number $13$ is squared it gives $169$. Then if we take the square root of $169$; $\sqrt{169}$ it gives $13$ and $-13$. Why is this so if we know that $13$ was positive and it was multiplied by itself and produced $169$.

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    $\begingroup$ Without additional information, if you are given $x^2 = 169$ then there are two possible solutions $\pm 13$. If you are given the additional fact that $x>0$ then this will disambiguate. $\endgroup$ – copper.hat Apr 24 '13 at 6:18
  • $\begingroup$ What if I'm given the additional fact that $x > 0$ as you said? I mean what do you mean by disambiguate here? $\endgroup$ – Samama Fahim Apr 24 '13 at 6:24
  • $\begingroup$ Well, if you are looking for a solution to $x^2=169$ then there are two possibilities as you have noted. So how do you choose? The problem at hand may have additional information that allows you to choose one of the two possibilities. For example, if you are trying to compute the width of a square that results in 169 square units, then the answer will be the positive one. $\endgroup$ – copper.hat Apr 24 '13 at 6:29
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Always note that $\sqrt{x}$ always gives a positive number.

So in your example: $$\sqrt{169} = +13$$

But, If an equation is given like: $$x^2 = 169$$ then, $$x = +13 (or) -13$$

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  • $\begingroup$ The ordinary North American high school convention has been that, for example, $\sqrt{256}=\pm 16$. When the students reach university, they need to become accustomed to the fact that in, say, a first year calculus course, $\sqrt{256}=16$. $\endgroup$ – André Nicolas Apr 24 '13 at 6:26
  • $\begingroup$ Oh ! But that is not correct ! May be the previous step would be something like $x^2 = 256$ $\endgroup$ – lsp Apr 24 '13 at 6:29
  • $\begingroup$ @AndréNicolas When I was in high school (which is quite recent) $\sqrt{x}$ always stood for the principal root. Maybe that's just particular to my high school. $\endgroup$ – EuYu Apr 24 '13 at 6:30
  • $\begingroup$ That's good, one less thing to unlearn. If your school was in Ontario, the Ontario curriculum has improved. $\endgroup$ – André Nicolas Apr 24 '13 at 6:33
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because squaring is not injective. Specifically, $(13)^2 = 169$ but $(-13)^2=169$ also. Square root is supposed to invert a square but in fact the inverse is not a function.

If you're talking in particular about what you put into a calculator, the calculator doesn't "remember" what the input value was. Internally, it stores whatever the result was. So your calculator doesn't know the difference between the result of $(-13)^2$ and $(13)^2$. There's no real reason that it couldn't, they just don't.

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Actual definition of square root function is $\sqrt{X^2}=|X|$

All square roots of $X^2$ is indeed $-\sqrt{X}$ and $\sqrt{X^2}$.

Although the principal square root of a positive number is only one of its two square roots, the designation "the square root" is often used to refer to the principal square root. For positive a, the principal square root can also be written in exponent notation, as $a^{\frac{1}{2}}$. Source:Square root

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  • $\begingroup$ Your first line is incorrect. $\sqrt{x^2} = |x|$ $\endgroup$ – EuYu Apr 24 '13 at 6:26
  • $\begingroup$ Right. I didn't consider the negatives at first. $\endgroup$ – Inceptio Apr 24 '13 at 6:31
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    $\begingroup$ A bit of nitpicking. You don't need the second $\sqrt{x^2} \neq -|x|$. Not only is it extraneous, it is wrong. Consider $x=0$. $\endgroup$ – EuYu Apr 24 '13 at 6:32
  • $\begingroup$ Thanks . I'm missing cases, I'm on anesthesia(laughs). $\endgroup$ – Inceptio Apr 24 '13 at 6:34

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