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 : $\pm y$
On
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.
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 $D$.