If x

Question

If  x < 0 , then  |x - \sqrt{(x - 1)^2}|  equals

A. 1

B. 1 - 2x

C. -2x - 1

D. 1 + 2x

E. 2x - 1

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chetan2u · Answer

Two ways..

(I) Simpler one....SUBSTITUTE for x... 
Let x=-3..
 |x - \sqrt{(x - 1)^2}| = |-3 - \sqrt{(-3 - 1)^2}| = |-3 - \sqrt{16}| = |-3-4|=7 
substitute x=-3 in choices..

A. 1....NO

B. 1 - 2x....1-(2*-3)=1+6=7.....YES

C. -2x - 1.....-(2*-3)-1=6-1=5...NO

D. 1 + 2x......1+(2*-3)=1-6=-5.....NO

E. 2x - 1.....(2*-3)-1=-6-1=-7...NO

B

(II) Algebraic method

As x<0...-x>0
 |x - \sqrt{(x - 1)^2}| = |x - \sqrt{(1-x)^2}| = |x-(1-x)|=|x-1+x|=|2x-1|=1-2x

B

Great question  Bunuel

GMATinsight · Answer

Substituting x = -1,  |x - \sqrt{(x - 1)^2}| = |-1 - \sqrt{(-1 - 1)^2}| = |-1-2| = 3

Checking Options

A. 1  OUT

B. 1 - 2x  Substituting x = -1, 1 - 2x = 1-2(-1) = 3 WORKS

C. -2x - 1  Substituting x = -1, -2x - 1 = 2-1 = 1 OUT

D. 1 + 2x  Substituting x = -1, 2x + 1 = 2(-1)+1 = -1 OUT

E. 2x - 1  Substituting x = -1, 2x - 1 = -2-1 = -3 OUT

Answer: Option B

Kritisood · Answer

If x < 0 , then |x - \sqrt{(x - 1)^2}| equals => |x-|x-1|| (x-1) would be <0 as x<0, therefore, when we remove the mod from (x-1) it will give us a negative ans => |x+x-1| => |2x-1| again (2x-1) would be negative always as x<0 (same logic as above) = 1-2x ans B

Archit3110 · Answer

given that x<0
so x is -ve value
let x=-3
we get 
 |x - \sqrt{(x - 1)^2}| 
l-3-√(-4)^2l
solve
=> 7
for given answer options we get 7 only at ; 1-2x
OPTION B

Onyedelmagnifico · Answer

Consider p>0 such that x=-p 
Then |x-√(x-1)²|=
|-p-√(-p-1)²|=
|-p-√((-1)(p+1))²|= 
|-p-√(-1)²(p+1)²|=
|-p-√(p+1)²|=
|-p-(p+1)|=|-p-p-1|= |-2p-1|=|(-1)(2p+1)|=|2p+1|=2p+1 (since p is positive)
Note p=-x 
Hence 
2p+1=-2x+1=1-2x

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Manku · Answer

We are given x<0
Keeping this constraint in mind, we will simplify the question stem.

|x - ((x - 1)^2)^(1/2)| becomes |x- |x - 1||

Since x<0 , |x-1| becomes -(x-1)

So we get , |x + x-1 | = |2x - 1|

Again , since x< 0 , |2x - 1| becomes -(2x-1) which is option B

Hope it helps

enricosanchez · Answer

If  x < 0 , then  |x - \sqrt{(x - 1)^2}|  equals

A. 1

B. 1 - 2x

C. -2x - 1

D. 1 + 2x

E. 2x - 1

A little bit tricky. I got ahead of myself... I saw the square root was squared, so I knew it was an absolute value, however on my paper didn't write that the entire equation as an absolute value.

This led me to -3 instead of 3. Which then gets you to trap answer 2x-1.

BrushMyQuant · Answer

↧↧↧ Detailed Video Solution to the Problem ↧↧↧

Given that \(x < 0\) and we need to find the value of \(|x - \sqrt{(x - 1)^2}|\)

Let x = -k
As x < 0
=> k > 0

=> \(|x - \sqrt{(x - 1)^2}|\) = \(|-k - \sqrt{(-k - 1)^2}|\)
= \(|-k - \sqrt{(k + 1)^2}|\)
= \(|-k - (k + 1)|\)
= \(|-k - k - 1|\)
= \(|-2k - 1|\)
= \(|-(2k + 1)|\)
= 2k + 1 (as 2k+ 1 > 0)

Now, x = -k
=> k = -x

=> 2k + 1 = -2x + 1 = 1 - 2x

So, Answer will be B
Hope it helps!

Watch the following video to learn How to Solve Absolute Value Problems

PalakFocus · Answer

hello, why cant square root of 16 be considered -4 , that way it would lead to 1 , how would you eliminate that path in the calculations?

Krunaal · Answer

On the GMAT, when a square root symbol is used , it always refers to the positive square root (or principal square root), and you should only consider the positive value. Here's link to the discussion: https://gmatclub.com/forum/sigh-denotes ... l#p3222939

bumpbot · Answer

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If x < 0, then |x - ((x - 1)^2)^(1/2)| equals

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