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As in gre, square root symbol means principal square root, what should be the answer in this case? Should we take only positive root i.e., $-y$ or both the roots $+$ or $-y$?

If $y < 0$, $\sqrt{-y|y|} =$ ?

A : $y$

B : $1$

C : $-1$

D : $-y$

E : $\pm y$

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2 Answers 2

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Recall that the absolute value is defined by $$ |x| = \begin{cases} x \text{ if } x\geq 0,\\ -x\text{ if } x<0. \end{cases} $$ In other words: when the number is positive, the absolute value is the same number; when the number is negative, you have tu change the sign of the number, that is why you have to add a minus. In the problem, you have to do the following substitution, $$ −y|y|=(−y)(−y)=(−y)^2 $$ because, since $y$ is negative, the absolute value $|y|$ is precisely $-y$.

Thus, we have to find the square root of $(−y)^2$, i.e., a positive nuber whose square is $(−y)^2$. But this is precisely $−y$, since $y<0$. Hence, the answer is $D$.

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  • $\begingroup$ I can't understand why this answer, the only one correct/complete/ explained, was downvoted. Perhaps I'm missing something, but in the meantime I shall upvote it. $\endgroup$
    – DonAntonio
    Sep 16, 2016 at 12:13
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    $\begingroup$ Recall that the absolute value is defined by $$ |x| = \begin{cases} x \text{ if } x\geq 0,\\ -x\text{ if } x<0. \end{cases} $$ In other words: when the number is positive, the absolute value is the same number; when the number is negative, you have tu change the sign of the number, that is why you have to add a minus. What I did in $-y|y| = (-y)^{2}$ was the following substitution: $$ -y|y| = (-y) (-y) = (-y)^{2} $$ because, since $y$ is negative, the absolute value $|y|$ is precisely $-y$. $\endgroup$
    – LaloVelasco
    Sep 16, 2016 at 12:40
  • $\begingroup$ I encourage anyone to upvote this answer since it indeed is better than the one I provided. Also @LaloVelasco could you edit the answer so to include the text of your comment here? That way the answer perhaps would be better and my vote would be ulocked so I could upvote this answer so to move it up on the thread. $\endgroup$
    – bonehead
    Sep 16, 2016 at 12:55
  • $\begingroup$ Thanks for your encouraging. I just edited my old answer. $\endgroup$
    – LaloVelasco
    Sep 16, 2016 at 13:10
  • $\begingroup$ @LaloVelasco thank you :) $\endgroup$
    – JKLM
    Sep 16, 2016 at 13:40
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No, $\sqrt{a}$ is a strictly defined single number, defined as

Given a real number $a$, the number $\sqrt{x}$ is the non-negative solution to the equation $x^2=a$

So, for example, $\sqrt9$ does not equal "plus or minus $3$", it equals $3$. It is true that the equation $x^2=9$ has two real solutions, $+$ and $-3$, but the task in front of you isn't asking about that.

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  • $\begingroup$ A is the answer? $\endgroup$
    – JKLM
    Sep 16, 2016 at 11:57
  • $\begingroup$ @Saurabh I don't know, do you think that $y$ is the non-negative solution to the equation $x^2=\sqrt{-y|y|}$? $\endgroup$
    – 5xum
    Sep 16, 2016 at 11:58
  • $\begingroup$ yes i think so..because square always give positive number, we can not take negative numbers squares. $\endgroup$
    – JKLM
    Sep 16, 2016 at 11:59
  • $\begingroup$ @Saurabh That's true, but is $y$ a non-negative solution? $\endgroup$
    – 5xum
    Sep 16, 2016 at 12:00
  • $\begingroup$ Don't forget to remember the initial conditions while thinking about the question above. $\endgroup$
    – bonehead
    Sep 16, 2016 at 12:03

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