by definition, absolute value function is:
|x| = x (for x >= 0)
|x| = -x (for x < 0)
so, [-x.|x|]^(1/2) = [-x.-x]^(1/2)
or [(-x)^2]^(1/2)
or -x.
Also, for a square root of any number to be defined, the number being square rooted must not be negative, thus the answer must also be >0.
But, x<0 is already given, hence, x can never be the answer.
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gmat740
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so, [-x.|x|]^(1/2) = [-x.-x]^(1/2)
or [(-x)^2]^(1/2)
or -x.
Also, for a square root of any number to be defined, the number being square rooted must not be negative, thus the answer must also be >0.
But, x<0 is already given, hence, x can never be the answer.
The number inside the square root is positive.And square root of a number can be positive or negative.
Eg: Sq-rt[(-2)(-2)]
Sqrt[4]
ans = +/- 2
ie: answer can be +2 or -2
Since x<0
So automatically the value will be negative so there won't be any requirement to put an additional (-) sign
Please give an explanation to my doubt
You are right, the square root can give +ve or -ve values...
but if in the equation you substitute |x| with -x as x<0, then you do not have to take the square root of the answer the indices cancel out.
((y)^2)^0.5 is y, not -y. since we know we are squaring y and then taking its square root.
but had we been given the numeric value of y^2 instead, say for example
(25)^0.5, then the answer could have been +/- 5
but if in the equation you substitute |x| with -x as x<0, then you do not have to take the square root of the answer the indices cancel out.
((y)^2)^0.5 is y, not -y. since we know we are squaring y and then taking its square root.
but had we been given the numeric value of y^2 instead, say for example
(25)^0.5, then the answer could have been +/- 5
Cheers,
Dubes
Dubes
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Vemuri
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Hi Kapsii, I agree with you that |x| for a negative x is = -x (the -ve variable value will make it a +ve x). But, when it comes to the expression, I am not sure how you derived -x.
[-x * -x]^(1/2) ==> [x^2]^(1/2) ==> x (after the indices cancel out). How did you derive -x?
[-x * -x]^(1/2) ==> [x^2]^(1/2) ==> x (after the indices cancel out). How did you derive -x?
well, where we differ is:
You say [-x * -x]^(1/2) ==> [x^2]^(1/2)
I say [(-x) *(-x)]^(1/2) ==> [(-x)^2]^(1/2)
i.e. -x * -x is more accurately described as (-x)^2 than x^2. Even though the two will give you same values, this question is slightly ambiguous, the correct answer obviously is +/- x, but if we have to choose between x & -x, I would choose -x as I have focused on keeping -x as a variable and not substituting the value of -x*-x as x^2.
You say [-x * -x]^(1/2) ==> [x^2]^(1/2)
I say [(-x) *(-x)]^(1/2) ==> [(-x)^2]^(1/2)
i.e. -x * -x is more accurately described as (-x)^2 than x^2. Even though the two will give you same values, this question is slightly ambiguous, the correct answer obviously is +/- x, but if we have to choose between x & -x, I would choose -x as I have focused on keeping -x as a variable and not substituting the value of -x*-x as x^2.
Cheers,
Dubes
Dubes
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Vemuri
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Yep, your explanation leaves no scope for argumentkapsii wrote:well, where we differ is:
You say [-x * -x]^(1/2) ==> [x^2]^(1/2)
I say [(-x) *(-x)]^(1/2) ==> [(-x)^2]^(1/2)
i.e. -x * -x is more accurately described as (-x)^2 than x^2. Even though the two will give you same values, this question is slightly ambiguous, the correct answer obviously is +/- x, but if we have to choose between x & -x, I would choose -x as I have focused on keeping -x as a variable and not substituting the value of -x*-x as x^2.
I wonder if we can think like this in the real test 
gmat740: What is the OA explanation?
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gmat740
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Well this is gmat prep question,I have only the OA and not the explanation.
But the answer seems to ambiguous and I am still not satisfied with replies posted.
Even if we have y=+y or y=-y, the value of y^2 or (-y)^2 will never change ie:
both will be positive y^2
and now when we take the sq-rt of( y^2) = +/- y
SO how one can say its +y or -y??
But the answer seems to ambiguous and I am still not satisfied with replies posted.
Take a look at the Bold Part((y)^2)^0.5 is y, not -y. since we know we are squaring y and then taking its square root.
but had we been given the numeric value of y^2 instead, say for example
(25)^0.5, then the answer could have been +/- 5
Even if we have y=+y or y=-y, the value of y^2 or (-y)^2 will never change ie:
both will be positive y^2
and now when we take the sq-rt of( y^2) = +/- y
SO how one can say its +y or -y??
We are given that X<0. This statement is important. Think of only numbers to the left of Zero on the number line. (-1, 1.5, -2, -3....). X has no number on the positive segment of the number line. Let x =-1gmat740 wrote:If x<0,
then [-x|x|]^0.5
a. -x
b. -1
c. 1
d. x
e. 0
My answer D
OA A
can someone explain?
|-1| = 1. Plugging -1 under the radical gives
(-(-1) 1)^.5 = 1^.5 = 1, -1. But if we look in our parenthesis above we don't see 1. So the answer is -1.
In theory (x^2)^.5 has two values. A posive number and a negative counerpart. Let's imagine 1, -1 as these two values when we take the squre root of 1. In the first case x=1, the positve value, and in the 2nd and -x=1, the second case. NOTE THAT -X CANNOT = -1 (-X=-1) B/C THAT WOULD STILL GIVE X=1, THE POSITIVE CONDITION AND WOULD NOT MEET THE SECOND CONDITION OF -1). So the 2nd condition has to be -x =1 or x=-1. x=-1 is the same as 1=-x if WE MUST PRODUCE -1 AS REQUIRED BY THE QUESTION!! Therein lies the rub of that question. Since we are told that our X can only be Negative, of the two choices coming out of radical, the only one that meets this negative condition is when -x=1 or when X=-1. That is why the answer out of the radical must be -x.
