Exercise from OFFICIAL GMAT PREP- I could use a little help

[bobdylan]

1) If X<0, then sq. root of -X . I X I is:

a) - X
b) - 1
c) 1
d) X
e) sq root of X

Thanks in advance!

[Anurag@Gurome]

1) If X<0, then sq. root of -X . I X I is:

a) - X
b) - 1
c) 1
d) X
e) sq root of X

Thanks in advance!

Note that √(x²) = |x|, which means that the square root cannot give a negative result.
So, √(x²) ≥ 0
Now if x = 5, then √(x²) = √25 = 5 = x = positive.
If x = -5, then √(x²) = √25 = 5 = -x = positive.

So, we can say that √(x²) = x if x ≥ 0.
and √(x²) = -x, if x < 0.

√(-x *|x|) = √(-x * -x) = √(x²) = |x| = -x

The correct answer is A.

[GMATGuruNY]

1) If X<0, then √( -X*IX| ) is:

a) - X
b) - 1
c) 1
d) X
e) √X

Thanks in advance!

Please note that √ means the POSITIVE ROOT ONLY.

Let x=-2.
Then √( -x*Ix| ) = √( -(-2)*I-2| ) = √4 = 2. This is our target.
Now we plug x=-2 into the answers to see which yields our target of 2.
Only answer choice A works:
-x = -(-2) = 2.

The correct answer is A.

[Shalabh's Quants]

1) If X<0, then sq. root of -X . I X I is:

a) - X
b) - 1
c) 1
d) X
e) sq root of X

Thanks in advance!

If I understood correctly...

It asks for sqrt [(-x)*|x|];

As X < 0; hence sqrt [(-x)*|x|] = sqrt[(x)*|-x|] = sqrt[(x)*x]= sqrt(x^2) = x. (With sign adjustment)

X is of opposite sign of -X, hence answer should be A.

------------

One approach can be take any random value of X = -4. You will get sqrt [(-x)*|x|] = 4 or -X.

[ranjeet75]

[quote="GMATGuruNY"][quote="bobdylan"]1) If X<0, then √( -X*IX| ) is:

a) - X
b) - 1
c) 1
d) X
e) √X

Thanks in advance![/quote]

Please note that √ means the POSITIVE ROOT ONLY.

Let x=-2.
Then √( -x*Ix| ) = √( -(-2)*I-2| ) = √4 = 2. This is our target.
Now we plug x=-2 into the answers to see which yields our target of 2.
Only answer choice [spoiler]A[/spoiler] works:
-x = -(-2) = 2.

The correct answer is [spoiler]A[/spoiler].[/quote]

If we take x = -2, then the answer is 2, so in the question it is given that x<0 so the answer should be x i.e.,

Option D should be correct.

Please help me where i am missing