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For x < 0. Simplify \(\sqrt{-(x + 1)*|x-1| + 1}\)?

A. x B. x - 1 C. x + 1 D. – x E. – x + 1

One can also use plug-in method for this problem.

Since given that \(x<0\), then say \(x=-1\), then \(\sqrt{-(x + 1)*|x-1| + 1}=\sqrt{-(-1 + 1)*|-1-1| + 1}=\sqrt{0+1}=1\).

Now, plug \(x=-1\) into the answer choices to see which one yields 1. Only answer choice D works: \(-x=-(-1)=1\)

Answer: D.

Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.

Re: For x < 0. Simplify {-(x + 1) |x-1| + 1}^1/2? [#permalink]

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11 Sep 2012, 22:41

Bunuel/Karishma

How can this be done using algebra. As explained by cyber we have taken value of !X-1! as negative, However we have only been provided with X<0. May be I am missing something..Plz Help

Re: For x < 0. Simplify {-(x + 1) |x-1| + 1}1/2? [#permalink]

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09 Oct 2012, 20:19

1

This post received KUDOS

cyberjadugar wrote:

Vamshi8411 wrote:

Q) For x < 0. Simplify {-(x + 1) |x-1| + 1}1/2?

A) x

B) x - 1

C) x + 1

D) – x

E) – x + 1

Hi,

for x<0, |x-1| = 1-x \(\sqrt{1-(x + 1) |x-1|}\) =\(\sqrt{1-(x + 1)(1-x)}\) =\(\sqrt{1-1+x^2}\) =\(\sqrt{x^2}\) =|x|, again x<0 = -x

Answer (D)

Regards,

Hi!

I'm confused as why the answer isn't simply x (as opposed to-x). The absolute value of x, when simplified, is positive x, even if it is stated that x<0. Why do you go back and add a negative?

Re: For x < 0. Simplify {-(x + 1) |x-1| + 1}1/2? [#permalink]

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09 Oct 2012, 23:56

1

This post received KUDOS

egiles wrote:

cyberjadugar wrote:

Vamshi8411 wrote:

Q) For x < 0. Simplify {-(x + 1) |x-1| + 1}1/2?

A) x

B) x - 1

C) x + 1

D) – x

E) – x + 1

Hi,

for x<0, |x-1| = 1-x \(\sqrt{1-(x + 1) |x-1|}\) =\(\sqrt{1-(x + 1)(1-x)}\) =\(\sqrt{1-1+x^2}\) =\(\sqrt{x^2}\) =|x|, again x<0 = -x

Answer (D)

Regards,

Hi!

I'm confused as why the answer isn't simply x (as opposed to-x). The absolute value of x, when simplified, is positive x, even if it is stated that x<0. Why do you go back and add a negative?

Thanks!

What is positive x when x is negative? What do you mean by absolute value of x, when simplified? Is \(|-7| = -7?\) NOOOO! \(|-7|=-(-7)=7>0!!!\) Absolute value is always non-negative.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: For x < 0. Simplify {-(x + 1) |x-1| + 1}1/2? [#permalink]

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10 Oct 2012, 09:02

Hi,

for x<0, |x-1| = 1-x \(\sqrt{1-(x + 1) |x-1|}\) =\(\sqrt{1-(x + 1)(1-x)}\) =\(\sqrt{1-1+x^2}\) =\(\sqrt{x^2}\) =|x|, again x<0 = -x

Answer (D)

Regards,[/quote]

Hi!

I'm confused as why the answer isn't simply x (as opposed to-x). The absolute value of x, when simplified, is positive x, even if it is stated that x<0. Why do you go back and add a negative?

Thanks![/quote]

What is positive x when x is negative? What do you mean by absolute value of x, when simplified? Is \(|-7| = -7?\) NOOOO! \(|-7|=-(-7)=7>0!!!\) Absolute value is always non-negative.[/quote]

Hi EvaJager,

First, many thanks for the help you give me and others on this board. It is much appreciated.

Here is where my answer differed from yours. When I solved it, I reached this point:

= |x| = x

Here is what you did:

= |x| = -x

I am confused why you said the absolute value of x is -x. I thought all the absolute value of all numbers is non-negative.

Re: For x < 0. Simplify {-(x + 1) |x-1| + 1}1/2? [#permalink]

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10 Oct 2012, 09:55

egiles wrote:

Hi,

for x<0, |x-1| = 1-x \(\sqrt{1-(x + 1) |x-1|}\) =\(\sqrt{1-(x + 1)(1-x)}\) =\(\sqrt{1-1+x^2}\) =\(\sqrt{x^2}\) =|x|, again x<0 = -x

Answer (D)

Regards,

Hi!

I'm confused as why the answer isn't simply x (as opposed to-x). The absolute value of x, when simplified, is positive x, even if it is stated that x<0. Why do you go back and add a negative?

Thanks![/quote]

-------------------------------- What is positive x when x is negative? What do you mean by absolute value of x, when simplified? Is \(|-7| = -7?\) NOOOO! \(|-7|=-(-7)=7>0!!!\) Absolute value is always non-negative.[/quote]

Hi EvaJager,

First, many thanks for the help you give me and others on this board. It is much appreciated.

Here is where my answer differed from yours. When I solved it, I reached this point:

= |x| = x

Here is what you did:

= |x| = -x

I am confused why you said the absolute value of x is -x. I thought all the absolute value of all numbers is non-negative.

Thanks again! Eric[/quote]

------------------------- I am asking the same question again: if \(x = -7,\) is \(|-7|=-7\)??? NO!!! \(|-7| = 7\). But \(x\) is not \(7, \,\,x\) is \(-7.\) What is the connection between \(-7\) and \(7?\) Simply, \(7 = -(-7).\) When \(x\) is negative, multiplying it by \(-1\) it turns it into a positive number. Therefore, \(|x|=-x\) for \(x<0.\) You cannot write \(|-7|=-7.\) A letter denoting a number if doesn't have a minus sign in front of it, it doesn't mean it cannot be negative. \(x\) doesn't automatically designate a positive number. You are stating yourself that \(x\) is negative!

Absolute value of a number expresses the distance on the number line between that number and 0. Distance between \(-7\) and \(0\) is \(7\). A number \(x\) can be negative, for example \(x=-7\). And \(-x\) can be positive, if \(x=-5\), because \(-(-5)=5.\)
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: For x < 0. Simplify {-(x + 1) |x-1| + 1}^1/2? [#permalink]

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07 Jan 2013, 08:18

Can someone please explain why |x-1| = 1-x. I solved the question by filling in -2 and got the answer, but I really want to understand the algebra? Thanks in advance!

Re: For x < 0. Simplify {-(x + 1) |x-1| + 1}^1/2? [#permalink]

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Re: For x < 0. Simplify {-(x + 1) |x-1| + 1}^1/2? [#permalink]

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Re: For x < 0. Simplify {-(x + 1) |x-1| + 1}^1/2? [#permalink]

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Re: For x < 0. Simplify {-(x + 1) |x-1| + 1}^1/2? [#permalink]

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03 Sep 2017, 17:26

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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