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Set 22 --Q.21 --square root of √(-x│x│)

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by prachich1987 » Mon Dec 13, 2010 5:53 am
If x < 0, then √[(-x)*(mod of x)] is


A. -x
B. -1
C. 1
D. x
E. √x

[spoiler]The OA is A, But I don't understand why it cannot be D.
I have read somewhere that in GMAT, there is no -ve square root.
We have to consider only +ve roots.
Plz explain[/spoiler]
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by shovan85 » Mon Dec 13, 2010 6:09 am
prachich1987 wrote:If x < 0, then √[(-x)*(mod of x)] is


A. -x
B. -1
C. 1
D. x
E. √x

[spoiler]The OA is A, But I don't understand why it cannot be D.
I have read somewhere that in GMAT, there is no -ve square root.
We have to consider only +ve roots.
Plz explain[/spoiler]
√[(-x)*(mod of x)]

= √[(-x) | x |]

What is |x|?

|x| = x when x > 0
|x| = -x when x < 0

Thats why the answer is -x as question explicitly mentions x is less than zero.

As per considering the postive roots be sure you are not dealing with Inequality or Equations or Absolute Values.
In these 3 cases you must consider both of the cases.

I agree you have read this but for cases as for example say you know in Geometry you know the area of the square then for determining side you cannot consider the -ve value.
Last edited by shovan85 on Mon Dec 13, 2010 6:13 am, edited 1 time in total.
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by prachich1987 » Mon Dec 13, 2010 6:13 am
shovan85 wrote:
prachich1987 wrote:If x < 0, then √[(-x)*(mod of x)] is


A. -x
B. -1
C. 1
D. x
E. √x

[spoiler]The OA is A, But I don't understand why it cannot be D.
I have read somewhere that in GMAT, there is no -ve square root.
We have to consider only +ve roots.
Plz explain[/spoiler]
√[(-x)*(mod of x)]

= √[(-x) | x |]

What is |x|?

|x| = x when x > 0
|x| = -x when x < 0

Thats why the answer is -x
Ok bt ultimately it becomes (-x)(-x)=sq of x
So the answer can be x or -x
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by shovan85 » Mon Dec 13, 2010 6:20 am
prachich1987 wrote: Ok bt ultimately it becomes (-x)(-x)=sq of x
So the answer can be x or -x
Yes. I know this confuses a lot. Try to visualize by considering a -ve value of x.

Take an example where x < 0

Say x = -2.

sqrt(x |x|) = sqrt[ (-2) * |-2| ]

Now tell me can you consider |-2| = 2? No because it will make sqrt[ (-2) * |-2| ] = sqrt(-4) which leads to Complex imaginary mathematics.
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by N:Dure » Mon Dec 13, 2010 8:51 am
shovan85 wrote:
prachich1987 wrote:If x < 0, then √[(-x)*(mod of x)] is


A. -x
B. -1
C. 1
D. x
E. √x

[spoiler]The OA is A, But I don't understand why it cannot be D.
I have read somewhere that in GMAT, there is no -ve square root.
We have to consider only +ve roots.
Plz explain[/spoiler]
√[(-x)*(mod of x)]

= √[(-x) | x |]

What is |x|?

|x| = x when x > 0
|x| = -x when x < 0

Thats why the answer is -x as question explicitly mentions x is less than zero.

As per considering the postive roots be sure you are not dealing with Inequality or Equations or Absolute Values.
In these 3 cases you must consider both of the cases.

I agree you have read this but for cases as for example say you know in Geometry you know the area of the square then for determining side you cannot consider the -ve value.
|x| = x when x > 0
|x| = -x when x < 0

I believe this is the key here, because when it's root x^2 then you'll pick the negative value.
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by mj78ind » Mon Dec 13, 2010 9:41 am
prachich1987 wrote:If x < 0, then √[(-x)*(mod of x)] is


A. -x
B. -1
C. 1
D. x
E. √x

[spoiler]The OA is A, But I don't understand why it cannot be D.
I have read somewhere that in GMAT, there is no -ve square root.
We have to consider only +ve roots.
Plz explain[/spoiler]
Another approach just picka negative number

Say x = -5, solving the equation gives sqrt{(-(-5))*(mod(-5)}
gives 5 which is -x

This saves time as well.......

Hope it helps!
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