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If x<0, then root(-x*|x|) is:

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New post 14 Jan 2015, 10:51
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Re: If x<0, then root(-x*|x|) is: [#permalink]

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New post 20 Jan 2016, 05:32
Hello from the GMAT Club BumpBot!

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Re: If x<0, then root(-x*|x|) is: [#permalink]

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New post 16 Apr 2016, 08:00
nss123 wrote:
If x<0, then \(\sqrt{-x|x|}\) is:

A. -x
B. -1
C. 1
D. x
E. \(\sqrt{x}\)


if x <0 then |x| = -x so this way we get \(\sqrt{(-x)*(-x)}\) i.e. \(\sqrt{x^2}\) i.e. |x|. As already said, |x| = -x, Option A is correct.
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Re: If x<0, then root(-x*|x|) is: [#permalink]

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New post 16 Apr 2016, 10:59
HKD1710 wrote:
nss123 wrote:
If x<0, then \(\sqrt{-x|x|}\) is:

A. -x
B. -1
C. 1
D. x
E. \(\sqrt{x}\)


if x <0 then |x| = -x so this way we get \(\sqrt{(-x)*(-x)}\) i.e. \(\sqrt{x^2}\) i.e. |x|. As already said, |x| = -x, Option A is correct.


This is incorrect. If x < 0, then we get \(\sqrt{- (-x)*(-x)}\) i.e. \(\sqrt{-x^2}\)

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Re: If x<0, then root(-x*|x|) is: [#permalink]

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New post 16 Apr 2016, 11:21
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atturhari wrote:
HKD1710 wrote:
nss123 wrote:
If x<0, then \(\sqrt{-x|x|}\) is:

A. -x
B. -1
C. 1
D. x
E. \(\sqrt{x}\)


if x <0 then |x| = -x so this way we get \(\sqrt{(-x)*(-x)}\) i.e. \(\sqrt{x^2}\) i.e. |x|. As already said, |x| = -x, Option A is correct.


This is incorrect. If x < 0, then we get \(\sqrt{- (-x)*(-x)}\) i.e. \(\sqrt{-x^2}\)


Hi atturhari,

\(\sqrt{-x^2}\) is NOT CORRECT. lets say x=-2. put this value in \(\sqrt{- (-x)*(-x)}\) and you get \(\sqrt{- (-(-2))*(-(-2))}\) i.e. \(\sqrt{- (2)*(2)}\) i.e. \(\sqrt{- 4}\) , which is not defined.

\(\sqrt{(-x)*(-x)}\) is correct. because x is negative and in this case first of all, you just place |x| in terms of x and i.e -x. Multuply -x with itself will give x^2. now square root of x^2 is nothing but |x| which is -x in our case.
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Re: If x<0, then root(-x*|x|) is: [#permalink]

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New post 16 Apr 2016, 11:41
atturhari wrote:
HKD1710 wrote:
nss123 wrote:
If x<0, then \(\sqrt{-x|x|}\) is:

A. -x
B. -1
C. 1
D. x
E. \(\sqrt{x}\)


if x <0 then |x| = -x so this way we get \(\sqrt{(-x)*(-x)}\) i.e. \(\sqrt{x^2}\) i.e. |x|. As already said, |x| = -x, Option A is correct.


This is incorrect. If x < 0, then we get \(\sqrt{- (-x)*(-x)}\) i.e. \(\sqrt{-x^2}\)


HKD1710 has already provided a good explanation but you need to realize that when you assume x<0 , only the |x| portion of the expression \(\sqrt{(-x)*|x|}\) will change to \(\sqrt{(-x)*(-x)}\) and NOT \(\sqrt{- (-x)*(-x)}\)

You are writing x as -x as well which is NOT correct, giving you an extra -1 than what is should be. 'x' will remain as such.

Hope this helps.
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Re: If x<0, then root(-x*|x|) is: [#permalink]

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New post 17 Apr 2017, 08:30
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Re: If x<0, then root(-x*|x|) is: [#permalink]

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New post 06 Aug 2017, 07:35
parker wrote:
Nice work whiplash! I agree 100% that plugging in numbers is often the most efficient (and least susceptible to careless errors) method on this kind of problem. One addt'l tip--it can be a timesaver to avoid 0, 1 (& -1), and numbers in the answer choices on a problem solving question (as opposed to on data sufficiency question, when you actually WANT to find the exceptions to the rule) because those are special numbers with special properties (the result of squaring 0 and 1 is the same as the number you put in, which is unusual) . Notice that the result of plugging in -1 is 1, which is choice A but it is *also* choice C. If you start with a small prime number like 2 or 3 (or in this case 2's negative relative, -2) you can sometimes save yourself a second pass.




hi

please comment on below...?

√- x^2

if the square root over and the square get cancelled out, we are left with "-x"

thanks in advance ...

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If x<0, then root(-x*|x|) is: [#permalink]

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New post 06 Aug 2017, 09:02
I am not sure if my solution makes sense, but that's how I get it into my mind(negative values under the square root)

\sqrt{−x∗|x|} = -(x*|x|)^1/2= -(x^2)^1/2 = - (x) = -x
Can someone comment please

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Re: If x<0, then root(-x*|x|) is: [#permalink]

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New post 06 Aug 2017, 09:26
cbh wrote:
I am not sure if my solution makes sense, but that's how I get it into my mind(negative values under the square root)

\sqrt{−x∗|x|} = -(x*|x|)^1/2= -(x^2)^1/2 = - (x) = -x
Can someone comment please



hey

I have done the same thing. Please see my post....

I am also seeking sage advice from experts ...

thanks
cheers :lol:

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Re: If x<0, then root(-x*|x|) is:   [#permalink] 06 Aug 2017, 09:26

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