If x

Question

If x < 0, then  \sqrt{-x|x|}  is:

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

Bunuel · Accepted Answer

SOME NOTES: 1. GMAT is dealing only with Real Numbers : Integers, Fractions and Irrational Numbers. 2. Any nonnegative real number has a unique non-negative square root called the principal square root and unless otherwise specified, the square root is generally taken to mean the principal square root . When the GMAT provides the square root sign for an even root, such as \sqrt{x} or \sqrt {x} , then the only accepted answer is the positive root . That is, \sqrt{25}=5 , NOT +5 or -5. In contrast, the equation x^2=25 has TWO solutions, +5 and -5. Even roots have only non-negative value on the GMAT. Odd roots will have the same sign as the base of the root. For example, \sqrt {125} =5 and \sqrt {-64} =-4 . 3. \sqrt{x^2}=|x| . The point here is that as square root function can not give negative result then \sqrt{some \ expression}\geq{0} . So \sqrt{x^2}\geq{0} . But what does \sqrt{x^2} equal to? Let's consider following examples: If x=5 --> \sqrt{x^2}=\sqrt{25}=5=x=positive ; If x=-5 --> \sqrt{x^2}=\sqrt{25}=5=-x=positive . So we got that: \sqrt{x^2}=x , if x\geq{0} ; \sqrt{x^2}=-x , if x<0 . What function does exactly the same thing? The absolute value function: |x|=x , if x\geq{0} and |x|=-x , if x<0 . That is why \sqrt{x^2}=|x| . BACK TO THE ORIGINAL QUESTION: 1. If x<0 , then \sqrt{-x*|x|} equals: A. -x B. -1 C. 1 D. x E. \sqrt{x} \sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x . Note that as x<0 then -x=-negative=positive , so \sqrt{-x*|x|}=-x=positive as it should be. Or just substitute the some negative x , let x=-5<0 --> \sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x . Answer: A. Hope it's clear.

Bunuel · Answer

1. If x<0, then \sqrt{-x*|x|} equals: A. -x B. -1 C. 1 D. x E. \sqrt{x} Remember: \sqrt{x^2}=|x| . The point here is that square root function can not give negative result : wich means that \sqrt{some \ expression}\geq{0} . So \sqrt{x^2}\geq{0} . But what does \sqrt{x^2} equal to? Let's consider following examples: If x=5 --> \sqrt{x^2}=\sqrt{25}=5=x=positive ; If x=-5 --> \sqrt{x^2}=\sqrt{25}=5=-x=positive . So we got that: \sqrt{x^2}=x , if x\geq{0} ; \sqrt{x^2}=-x , if x<0 . What function does exactly the same thing? The absolute value function! That is why \sqrt{x^2}=|x| Back to the original question: \sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x Or just substitute the value let x=-5<0 --> \sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x . Answer: A. Hope it's clear.

Bunuel · Answer

Given:  x<0  Question:  y=\sqrt{-x*|x|} ?

Remember:  \sqrt{x^2}=|x| .

As  x<0 , then  |x|=-x  -->  \sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x .

Answer: A.

whiplash2411 · Answer

The easiest way for you to solve this problem would be to plug in a number and see what happens.

Let's say  x = -1

\sqrt{-x|x|}=\sqrt{-(-1)|-1|}=\sqrt{(1)(1)}=1 = -(-1)

So, your answer is A, -x.

hailtothethief23 · Answer

It is actually A.

suppose x is -2 then you have sqrt(2*2) = sqrt(4) = 2 = -x

note that x < 0 as otherwise the function does not exist.

shammokando · Answer

Hey I'm sorry guys, this still does not make sense. Everyone's argument here is that the square root of 4 is 2, that is just not true! The square root of 4 is 2 OR -2. We're just accustomed to thinking that 2 is the "standard root" but -2 is just as correct. Therefore the square of -2 (which is x in this case) is 4, and the squareroot of that is 2 OR -2! So it could be x or -x.

This seems wrong and no one's explanation makes any sense.

This seems wrong and no one's explanation makes any sense.

Bunuel · Answer

Red part is not correct.

THEORY:

GMAT is dealing only with   Real Numbers  : Integers, Fractions and Irrational Numbers.

When the GMAT provides the square root sign for an even root, such as  \sqrt{x}  or  \sqrt {x} , then the  only accepted answer is the positive root .

That is,  \sqrt{25}=5 , NOT +5 or -5. In contrast, the equation  x^2=25  has TWO solutions, +5 and -5.  Even roots have only a positive value on the GMAT.

Odd roots will have the same sign as the base of the root. For example,  \sqrt {125} =5  and  \sqrt {-64} =-4 .

Solution for the original question:

Given:  x<0  Question:  \sqrt{-x*|x|}=? .

Remember:  \sqrt{x^2}=|x| .

As  x<0 , then  |x|=-x  -->  \sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x .

Answer: A.

Hope it helps.

Bunuel · Answer

GMAT is dealing only with   Real Numbers  : Integers, Fractions and Irrational Numbers. So you won't see any question involving imaginary numbers.

This question also does not involve imaginary numbers as expression under the square root is non-negative (actually it's positive): we have  \sqrt{-x*|x|}  --> as  x<0  then  -x=positive  and  |x|=positive , so  \sqrt{-x*|x|}=\sqrt{positive*positive}=\sqrt{positive} .

Hope it's clear.

saxenashobhit · Answer

I had same question today

\sqrt{4}  = + or - 2. So answer should be + or - x. I don't get how can OA be -x

Bunuel · Answer

About the odd roots: odd roots will have the same sign as the base of the root. For example, \sqrt {125} =5 and \sqrt {-64} =-4 . So, if given that x<0 then |x|=-x and -x*|x|=(-x)*(-x)=positive*positive=x^2 , thus odd root from positive x^2 will be positive. But \sqrt {-x*|x}| will equal neither to x nor to -x: \sqrt {-x*|x|}=\sqrt {x^2}=x^{\frac{2}{3}} , for example if x=-8<0 then \sqrt {-x*|x|}=\sqrt {x^2}=\sqrt {64}=4 . If it were x<0 and \sqrt {-x^2*|x}|=? , then \sqrt {-x^2*|x}|=\sqrt {-x^2*(-x)}=\sqrt {x^3}=x<0 . Or substitute the value let x=-5<0 --> \sqrt {-x^2*|x}|=\sqrt {-25*5}=\sqrt {-125}=-5=x<0 . Now, back to the original question: If x<0, then \sqrt{-x*|x|} equals: A. -x B. -1 C. 1 D. x E. \sqrt{x} As square root function cannot give negative result , then options -1 (B) and x (D) can not be the answers as they are negative. Also \sqrt{x} (E) can not be the answer as even root from negative number is undefined for GMAT. 1 (C) also can not be the answer as for different values of x the answer will be different, so it can not be some specific value. So we are left with A. Now, if it were x>0 instead of x<0 then the question would be flawed as in this case the expression under the even root would be negative (-x*|x|=negative*positive=negative). Hope it's clear.

IanStewart · Answer

If x < 0, then |x| = -x. So by substituting, we have:

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

Now it's important to understand that √(x^2) is not necessarily equal to x. That is only true when x is positive (or zero). You can see, if you plug in any negative number here, say x = -3, that √(x^2) = √9 = 3, which is not equal to x because the sign changed; it's actually equal to -x. In general, √(x^2) is always equal to |x|. Since x < 0 in this question, √(x^2) = |x| = -x.

EMPOWERgmatRichC · Answer

Hi jchae90,

This question is perfect for TESTing VALUES.

We're told that X < 0, so let's TEST X = -2

We're asked to determine the value of..... √((-x)·|x|)

√((-(-2))·|-2|) = √(2)·|2|) = √4 = 2

So we're looking for an answer that equals 2 when X = -2

Answer A: –X = -(-2) = 2 This IS a match 
Answer B: -1 NOT a match
Answer C: 1 NOT a match
Answer D: X = -2 NOT a match
Answer E: √X = √2 NOT a match

Final Answer:  A

GMAT assassins aren't born, they're made,
Rich

JeffTargetTestPrep · Answer

Since x is less than zero, |x| = -x.

Thus, we have:

√(-x|x|) = √(-x(-x)) = √(x^2)

Recall that √(x^2) = |x|; however, |x| = -x, since x is less than zero, so we have:

√(-x|x|) = √(-x(-x)) = √(x^2) = |x| = -x

Answer: A

CrackverbalGMAT · Answer

Let's take x= -5
=>So √ (-x |x|) = √ (-(-5)|-5|) 
=> √ (5*5) = √ (25)
= 5 
= -(-5) 
= -x 
(option a)

Kimberly77 · Answer

Hi  brunel , Shouldn't −(−5) become +5?
Thanks for great answer always and your time in advanced.

Bunuel · Answer

Yes, -(-5) = 5, which is -x (x = -5).

BrushMyQuant · Answer

Given that x < 0 and we need to find the value of  \sqrt{-x|x|}

As x < 0, so assume x = -k where k is a positive number

=>  \sqrt{-x|x|}  =  \sqrt{-(-k)|-k|} 
=  \sqrt{k*k}  (As |Positive Number| = Positive Number)
=  \sqrt{k^2}  = k = -x

So,  Answer will be A .
Hope it helps!

Watch the following video to learn the Basics of Absolute Values

https://www.youtube.com/watch?v=huFiBk8PaGY

NEYR0N · Answer

Bunuel ,

Since we know x < 0

Would it be correct to do:

the   becomes    
 |x| will be +x since the absolute value is always positive

So we also get to sqrt(x^2)

ManifestDreamMBA · Answer

Given x < 0, |x|=-x

\sqrt{-x|x|}  
=  \sqrt{|x||x|}  
= |x|
=-x

If x < 0, then (-x*|x|)^(1/2) is:

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