If x<0, then sq.rt.(-x|x|) is
1. -x
2. -1
3. 1
4. X
5. Sq.rt(x)
[spoiler]OA:A[/spoiler]
PS: tricky sq rt
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x is negative, plug x = -x
sqrt[(-(-x) |-x|)]
=sqrt[(x*x)]
= sqrt[x^2]
= x
but its given that x is -ve
Therefore value = -x
Option A.
sqrt[(-(-x) |-x|)]
=sqrt[(x*x)]
= sqrt[x^2]
= x
but its given that x is -ve
Therefore value = -x
Option A.
Last edited by beat_gmat_09 on Mon Nov 29, 2010 5:52 am, edited 1 time in total.
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Something still eludes me on this question.
Say x= -4. Multiplying -x by |x|, I get 16. Doesn't this have two square roots?
Say x= -4. Multiplying -x by |x|, I get 16. Doesn't this have two square roots?
I'm really old, but I'll never be too old to become more educated.
Good question. I never actually explicitly thought about that one before, so I Googled it. The answer is that, while there are 2 square roots as you've stated (+ and -), the √ notation supposedly designates the "principle square root", which is the *positive* answer.
https://www.mathsisfun.com/algebra/square-root.html
That being said, this looks like a candidate question to be thrown out due to ambiguity. They really should not have offered option 4 to be absolutely clear.
https://www.mathsisfun.com/algebra/square-root.html
That being said, this looks like a candidate question to be thrown out due to ambiguity. They really should not have offered option 4 to be absolutely clear.
tomada wrote:Something still eludes me on this question.
Say x= -4. Multiplying -x by |x|, I get 16. Doesn't this have two square roots?
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Now ,x<0.haidgmat wrote:If x<0, then sq.rt.(-x|x|) is
1. -x
2. -1
3. 1
4. X
5. Sq.rt(x)
[spoiler]OA:A[/spoiler]
So, |x|=-x.
Hence,the ques. becomes sqrt[(-x)*(-x)]
=sqrt[x^2]
=|x|
Now since x<0,|x| would be -x which is A.
Alternate Solution:-Pick any -ve number(say -2) and put in the question.
Trick:-A person can confuse between option (1) and (4).Do keep in mind that you are dealing with a -ve number hence "X" can never be the answer.
It takes time and effort to explain, so if my comment helped you please press Thanks button
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