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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
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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!
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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.
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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.

Thanks again!
Eric
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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 \(-(-5)=5.\)
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\(\sqrt{-(x + 1)*|x-1| + 1}\)

Since x < 0, what is |x+1|?
So let x = -1, x-1 = -2
let x = -2, x-1 = -3
Since x-1 is always (-), |x-1| = -(x-1).

Transform the equation:
\(\sqrt{-(x+1)*-(x+1)+1}\)
\(\sqrt{x^2-1+1}\)
\(\sqrt{x^2}=|x|\)

Since x<0, what is |x|?
|x| = -(x)

Answer: D
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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!
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Bunuel
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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.

Hope it helps.

Hi

Hypothetically if we land up with a solution as \(\sqrt{- 4}\) what would be the answer.
Trying to understand the concept.

Thanks in advance.
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Vamshi8411
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.

Hope it helps.

Hi

Hypothetically if we land up with a solution as \(\sqrt{- 4}\) what would be the answer.
Trying to understand the concept.

Thanks in advance.

It would mean that that you've made an error in arithmetic because even roots, such as square roots from negative numbers are not defined on the GMAT. All numbers in the GMAT are by default real numbers.
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Vamshi8411
For x < 0. Simplify \(\sqrt{-(x + 1)*|x-1| + 1}\)?

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

Finally :dazed after getting this logic wrong twice, i got it right in this question

Keyword x < 0

\(\sqrt{-(x + 1)*|x-1| + 1}\)

\(\sqrt{-(x + 1) * - (x-1) + 1}\)

\(\sqrt{x^2}\)

\(x^2\) = |x|

x < 0

Value is -x
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Vamshi8411
For x < 0. Simplify \(\sqrt{-(x + 1)*|x-1| + 1}\)?

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

Responding to a pm:

Here is how you can simplify it algebraically:

|x - 1| = (x - 1) when x >= 1
|x - 1| = -(x - 1) when x < 1

Here, we know that x < 0 so x MUST be less than 1 too. Hence, |x - 1| = -(x - 1)

\(\sqrt{-(x + 1)*|x-1| + 1} = \sqrt{-(x + 1)*-(x - 1) + 1}\)
\(= \sqrt{(x^2 - 1) + 1}\)
\(= \sqrt{x^2}\)
= |x|

Now, since we know that x < 0, |x| = -x

Answer (D)
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Vamshi8411
For x < 0. Simplify \(\sqrt{-(x + 1)*|x-1| + 1}\)?

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

Set X=-2

So:

(-(-2+1)*|-2-1|+1)^.5 =

(1*3+1)^.5 = 2 = -X

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