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Why does this happen??

$$ y = \sqrt9 \implies y=3$$

$$ y^2 = 9 \implies y=+3,-3 $$

While both equations are in same sense.

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  • $\begingroup$ Because the square function is not injective. $\endgroup$ – Stefan Perko May 22 '16 at 9:38
  • $\begingroup$ The square root is not the inverse of the square function. Not that the square function could possibly have an inverse function at all... a function is not just the "rule" or "equation" that transform each element into something else, you must also consider WHICH elements it transforms, that is, its domain. $\endgroup$ – Dleep May 22 '16 at 9:39
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    $\begingroup$ This question is a duplicate of approximately $\sqrt{10000}$ older questions. $\endgroup$ – barak manos May 22 '16 at 9:40
  • $\begingroup$ @barakmanos.... If I cant ask my doubt differently what is the point of having such a huge community? and also my doubt was not cleared in previous questions, so i asked differently. $\endgroup$ – prog_SAHIL May 22 '16 at 10:01
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    $\begingroup$ We usually consider $\sqrt{x}$ to be the principal square root of $x$ which is always positive. You may use $\pm \sqrt{x}$ to indicate that you meant both of them. $\endgroup$ – TheRandomGuy May 22 '16 at 10:06
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By definition $\sqrt 9$ does not mean "any number whose square is $9$", but specifically "the non-negative number whose square is $9$".

So the equations $y=\sqrt 9$ and $y^2=9$ are not (as you assume) the same.

Rather, $y=\sqrt 9$ is the same as $y^2=9 \land y\ge 0$. It should not be surprising that ignoring the $y\ge 9$ condition will result in more solutions.

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  • $\begingroup$ This means if I square the first equation, It will result in more roots. $\endgroup$ – prog_SAHIL May 22 '16 at 10:03
  • $\begingroup$ @prog_SAHIL: Correct. Just like if you square the false claim $-4=4$ you will produce something true. $\endgroup$ – hmakholm left over Monica May 22 '16 at 10:04
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Hint:

Any positive real number $x>0$ has two opposite square roots defined as the numbers $y$ such that $y^2=x$, but the symbol $\sqrt{x}$, by definition indicate only the positive root. So : $\sqrt{y^2}=|y|.$

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Though not much technical in the second equation, both the values of y satisfy the equation while in the first -3 doesn't, though the statements are equivalent, we only take numbers that are applicable, or are satisfiable, though even in the first statement we could have a solution if we bring complex numbers up.

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You can define square root as positive root of equation $x^2 = a$. We know that there are exactly two such roots $x_0$ and $-x_0$, so the positive one we will denote $\sqrt a$ and the negative one $-\sqrt a$.

Thus, $\sqrt 9$ is by above definition equal to $3$, while the equation $x^2 = 9$ has two solutions, $\sqrt 9 = 3$ and $-\sqrt 9 = -3$.

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