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Intern  S
Joined: 10 Jul 2016
Posts: 44
If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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Question Stats: 48% (01:25) correct 52% (01:41) wrong based on 165 sessions

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If x<0, what is the value of $$\sqrt{(x-3)^4} + \sqrt{-x|x|}$$

A. -3
B. x+3
C. 3-2x
D. 2x-3
E. 2
GMAT Tutor G
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Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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6
Top Contributor
2
Whenever we have absolute value question and the variable is given as negative (like x<0 in this case) then the easiest way to solve the problem is to
take x=-k (where k is positive), this will make opening absolute values easy

[(x-3)^4]^1/4 + (-x|x|)^1/2
= [((-k)-3)^4]^1/4 + (-(-k)|-k|)^1/2
= [(-(k+3))^4]^1/4 + ((k)|-k|)^1/2
= [(k+3)^4]^1/4 + ((k)(k))^1/2 [as k is positive so |-k| will be equal to k]
= (k+3) + (k^2)^1/2
= k+3 + k = 2k + 3
[x = -k => k = -x]
= 2(-x) + 3 = 3-2x
Hope it helps!

siddharthsinha123 wrote:
If x<0

What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2

A. -3
B. x+3
C. 3-2x
D. 2x-3
E. 2

_________________
##### General Discussion
Math Expert V
Joined: 02 Aug 2009
Posts: 8586
Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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6
siddharthsinha123 wrote:
If x<0

What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2

A. -3
B. x+3
C. 3-2x
D. 2x-3
E. 2

Hi

Best way is to substitute a value for X..
Let x be -2..
So [(-2-3)^4]^1/4+(-(-2)*|-2|)^1/2= ((-5)^4)^1/4+4^1/2=5+2=7..
Substitute in choices
. A.-3. ......NO
B. x+3. .....-2+3=1...NO
C. 3-2x. .....3-(2*-2)=3+4=7.. YES
D. 2x-3. .....2*-2-3=-7......NO
E. 2. .....NO

C
_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 64073
Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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1
siddharthsinha123 wrote:
If x<0, what is the value of $$\sqrt{(x-3)^4} + \sqrt{-x|x|}$$

A. -3
B. x+3
C. 3-2x
D. 2x-3
E. 2

Formatted the question.
_________________
Intern  S
Joined: 26 Apr 2016
Posts: 20
Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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1
chetan2u wrote:
siddharthsinha123 wrote:
If x<0

What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2

A. -3
B. x+3
C. 3-2x
D. 2x-3
E. 2

Hi

Best way is to substitute a value for X..
Let x be -2..
So [(-2-3)^4]^1/4+(-(-2)*|-2|)^1/2= ((-5)^4)^1/4+4^1/2=5+2=7..
Substitute in choices
. A.-3. ......NO
B. x+3. .....-2+3=1...NO
C. 3-2x. .....3-(2*-2)=3+4=7.. YES
D. 2x-3. .....2*-2-3=-7......NO
E. 2. .....NO

C

Hi,

Let me understand that why [(-2-3)^4]^1/4 = 5. I think that [(-5)^4]^1/4 should be -5
Current Student G
Joined: 18 Jun 2016
Posts: 251
Location: United States (NY)
GMAT 1: 720 Q50 V38
GMAT 2: 750 Q49 V42
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WE: General Management (Other)
If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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1
abhishekmayank wrote:
chetan2u wrote:
siddharthsinha123 wrote:
If x<0

What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2

A. -3
B. x+3
C. 3-2x
D. 2x-3
E. 2

Hi

Best way is to substitute a value for X..
Let x be -2..
So [(-2-3)^4]^1/4+(-(-2)*|-2|)^1/2= ((-5)^4)^1/4+4^1/2=5+2=7..
Substitute in choices
. A.-3. ......NO
B. x+3. .....-2+3=1...NO
C. 3-2x. .....3-(2*-2)=3+4=7.. YES
D. 2x-3. .....2*-2-3=-7......NO
E. 2. .....NO

C

Hi,

Let me understand that why [(-2-3)^4]^1/4 = 5. I think that [(-5)^4]^1/4 should be -5

There are 2 rules Here.

1. General Rule of Thumb - Every Number, Positive or Negative, raised to an Even Power will give a Positive Number.

Here is the Mathematics of same:

$$(-10)^2 = (-10)*(-10) = (-1)*(-1)*(10)*(10) = (1) * (100) = 100$$

2. Rule followed in GMAT

If $$x^2 = 16$$
x = -4 or +4

But $$\sqrt{16} = +4$$
Intern  B
Joined: 29 Apr 2017
Posts: 16
Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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umg wrote:
abhishekmayank wrote:
chetan2u wrote:
[quote="siddharthsinha123"]If x<0

What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2

A. -3
B. x+3
C. 3-2x
D. 2x-3
E. 2

Hi

Best way is to substitute a value for X..
Let x be -2..
So [(-2-3)^4]^1/4+(-(-2)*|-2|)^1/2= ((-5)^4)^1/4+4^1/2=5+2=7..
Substitute in choices
. A.-3. ......NO
B. x+3. .....-2+3=1...NO
C. 3-2x. .....3-(2*-2)=3+4=7.. YES
D. 2x-3. .....2*-2-3=-7......NO
E. 2. .....NO

C

Hi,

Let me understand that why [(-2-3)^4]^1/4 = 5. I think that [(-5)^4]^1/4 should be -5

There are 2 rules Here.

1. General Rule of Thumb - Every Number, Positive or Negative, raised to an Even Power will give a Positive Number.

Here is the Mathematics of same:

$$(-10)^2 = (-10)*(-10) = (-1)*(-1)*(10)*(10) = (1) * (100) = 100$$

2. Rule followed in GMAT

If $$x^2 = 16$$
x = -4 or +4

But $$\sqrt{16} = +4$$[/quote]
But still as it is -5^4*1/4
Simplifying it will give -5^1 = -5

I am so confused Sent from my SM-N9200 using GMAT Club Forum mobile app
Current Student G
Joined: 18 Jun 2016
Posts: 251
Location: United States (NY)
GMAT 1: 720 Q50 V38
GMAT 2: 750 Q49 V42
GPA: 4
WE: General Management (Other)
Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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khushbumodi wrote:
umg wrote:
abhishekmayank wrote:

Hi,

Let me understand that why [(-2-3)^4]^1/4 = 5. I think that [(-5)^4]^1/4 should be -5

There are 2 rules Here.

1. General Rule of Thumb - Every Number, Positive or Negative, raised to an Even Power will give a Positive Number.

Here is the Mathematics of same:

$$(-10)^2 = (-10)*(-10) = (-1)*(-1)*(10)*(10) = (1) * (100) = 100$$

2. Rule followed in GMAT

If $$x^2 = 16$$
x = -4 or +4

But $$\sqrt{16} = +4$$

But still as it is -5^4*1/4
Simplifying it will give -5^1 = -5

I am so confused Hmm.. That is a valid point. I did not think of that. What I did was I solved the inside bracket first, applied the power 4 and took the 4th root. I don't know why your method is wrong though mine IS the right way to solve because otherwise we would have 2 different solutions of an operation.

I believe that one of the lesser known fallacies of Mathematics is at work here. For example, Here is One of those..

$$\sqrt{x} = \sqrt{(x)*(-1)*(-1)} = \sqrt{x}*\sqrt{(-1)}*\sqrt{(-1)} = - \sqrt{x}$$
but unless x=0, the above result cannot be true; however since we solved for a general term x, this must be universally true.

The fallacy lies in an operation that we did..

We cannot split $$\sqrt{(-1)*(-1)}$$ into $$\sqrt{(-1)}*\sqrt{(-1)}$$

P.S. If you don't know, $$(\sqrt{(-1)})^2 = (-1)$$.
Math Expert V
Joined: 02 Sep 2009
Posts: 64073
Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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1
2
siddharthsinha123 wrote:
If x<0, what is the value of $$\sqrt{(x-3)^4} + \sqrt{-x|x|}$$

A. -3
B. x+3
C. 3-2x
D. 2x-3
E. 2

Two things to remember:

(1) $$\sqrt{x^2}=|x|$$

(2) When $$x \le 0$$ then $$|x|=-x$$, or more generally when $$\text{some expression} \le 0$$ then $$|\text{some expression}| = -(\text{some expression})$$. For example: $$|-5|=5=-(-5)$$;

When $$x \ge 0$$ then $$|x|=x$$, or more generally when $$\text{some expression} \ge 0$$ then $$|\text{some expression}| = \text{some expression}$$. For example: $$|5|=5$$.

Back to the question:

According to (1) $$\sqrt{(x-3)^4}=|x-3|$$. Now, since it's given that x is negative, then x - 3 will also be negative, thus according to (2) $$|x-3| = -(x-3)=3-x$$.

Next, again since x is negative, then $$|x|=-x$$. Thus, $$\sqrt{-x|x|}=\sqrt{(-x)(-x)}=\sqrt{x^2}=|x|=-x$$.

So, finally we have $$\sqrt{(x-3)^4} + \sqrt{-x|x|}=3-x-x=3-2x$$.

_________________
Manager  B
Joined: 16 Jan 2017
Posts: 59
GMAT 1: 620 Q46 V29
Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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chetan2u wrote:
siddharthsinha123 wrote:
If x<0

What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2

A. -3
B. x+3
C. 3-2x
D. 2x-3
E. 2

Hi

Best way is to substitute a value for X..
Let x be -2..
So [(-2-3)^4]^1/4+(-(-2)*|-2|)^1/2= ((-5)^4)^1/4+4^1/2=5+2=7..
Substitute in choices
. A.-3. ......NO
B. x+3. .....-2+3=1...NO
C. 3-2x. .....3-(2*-2)=3+4=7.. YES
D. 2x-3. .....2*-2-3=-7......NO
E. 2. .....NO

C

I am confused. How can you substitute x for a negative number, when x<0? wouldn't x have to be positive?
Director  G
Joined: 02 Sep 2016
Posts: 625
Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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Bunuel wrote:
siddharthsinha123 wrote:
If x<0, what is the value of $$\sqrt{(x-3)^4} + \sqrt{-x|x|}$$

A. -3
B. x+3
C. 3-2x
D. 2x-3
E. 2

Two things to remember:

(1) $$\sqrt{x^2}=|x|$$

(2) When $$x \le 0$$ then $$|x|=-x$$, or more generally when $$\text{some expression} \le 0$$ then $$|\text{some expression}| = -(\text{some expression})$$. For example: $$|-5|=5=-(-5)$$;

When $$x \ge 0$$ then $$|x|=x$$, or more generally when $$\text{some expression} \ge 0$$ then $$|\text{some expression}| = \text{some expression}$$. For example: $$|5|=5$$.

Back to the question:

According to (1) $$\sqrt{(x-3)^4}=|x-3|$$. Now, since it's given that x is negative, then x - 3 will also be negative, thus according to (2) $$|x-3| = -(x-3)=3-x$$.

Next, again since x is negative, then $$|x|=-x$$. Thus, $$\sqrt{-x|x|}=\sqrt{(-x)(-x)}=\sqrt{x^2}=|x|=-x$$.

So, finally we have $$\sqrt{(x-3)^4} + \sqrt{-x|x|}=3-x-x=3-2x$$.

Hello Bunuel
When it's written |x|, don't we always take the positive value no matter what is inside. Why are we even considering the negative value of x when absolute value of a number is positive. Please explain.
Director  G
Joined: 02 Sep 2016
Posts: 625
Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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Bunuel, abhimahna, BrushMyQuant

Absolute value rule is really creating confusion here. I read that absolute value of an integer/expression is always positive no matter what the sign is inside |x|. Then can't we just write here:

-(x-3)+ sq. root of (-x*x)
=-x+3+ sq. root of (-(x)^2)
= -x+3+(-(x)^2)^1/2
=-x+3+|-1|*x
= -x+3+x
=3

When I am removing | |, I am just considering positive value of whatever is inside | |.

GMAT Tutor G
Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 629
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)
Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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Top Contributor
Hi Shiv,

You did everything right apart from opening the second part of the equation
sq. root of (-x*|x|)
Since x is negative so |x| = -x (since x is negative so -x will be positive)
sq. root of (-x*|x|) = sq. root of (-x*-x) = sq. root of (x^2)
Now whatever comes out of square root is always positive
So sq. root of (x^2) = |x|
Again, x is negative so |x| = -x (since x is negative so -x will be positive)
-x + 3 - x = 3-2x

Hope it helps!

Shiv2016 wrote:
Bunuel, abhimahna, BrushMyQuant

Absolute value rule is really creating confusion here. I read that absolute value of an integer/expression is always positive no matter what the sign is inside |x|. Then can't we just write here:

-(x-3)+ sq. root of (-x*x)
=-x+3+ sq. root of (-(x)^2)
= -x+3+(-(x)^2)^1/2
=-x+3+|-1|*x
= -x+3+x
=3

When I am removing | |, I am just considering positive value of whatever is inside | |.

_________________
Director  G
Joined: 02 Sep 2016
Posts: 625
Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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BrushMyQuant wrote:
Hi Shiv,

You did everything right apart from opening the second part of the equation
sq. root of (-x*|x|)
Since x is negative so |x| = -x (since x is negative so -x will be positive)
sq. root of (-x*|x|) = sq. root of (-x*-x) = sq. root of (x^2)
Now whatever comes out of square root is always positive
So sq. root of (x^2) = |x|
Again, x is negative so |x| = -x (since x is negative so -x will be positive)
-x + 3 - x = 3-2x

Hope it helps!

Shiv2016 wrote:
Bunuel, abhimahna, BrushMyQuant

Absolute value rule is really creating confusion here. I read that absolute value of an integer/expression is always positive no matter what the sign is inside |x|. Then can't we just write here:

-(x-3)+ sq. root of (-x*x)
=-x+3+ sq. root of (-(x)^2)
= -x+3+(-(x)^2)^1/2
=-x+3+|-1|*x
= -x+3+x
=3

When I am removing | |, I am just considering positive value of whatever is inside | |.

May I ask a very basic question here?

We sat that absolute value of any number is positive. So shouldn't we write only positive value after removing the bars.
i.e. |x|= x

It's a very basic question but because of this only I am making mistakes in absolute value questions.
GMAT Tutor G
Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 629
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)
Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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Top Contributor
|x| = x if x is positive or zero
= -x if x is negative or zero

Example
|-3| = 3 = -(-3) (so |x| = -x if x is negative)

Sometimes, it becomes complex to solve questions when x is given as negative. So, to make things simple I suggest to take x = -k (where k is positive) and substitute x=-k in all the equations. Check my first reply in this thread to understand it better.

Hope it helps!

Shiv2016 wrote:
BrushMyQuant wrote:
Hi Shiv,

You did everything right apart from opening the second part of the equation
sq. root of (-x*|x|)
Since x is negative so |x| = -x (since x is negative so -x will be positive)
sq. root of (-x*|x|) = sq. root of (-x*-x) = sq. root of (x^2)
Now whatever comes out of square root is always positive
So sq. root of (x^2) = |x|
Again, x is negative so |x| = -x (since x is negative so -x will be positive)
-x + 3 - x = 3-2x

Hope it helps!

Shiv2016 wrote:
Bunuel, abhimahna, BrushMyQuant

Absolute value rule is really creating confusion here. I read that absolute value of an integer/expression is always positive no matter what the sign is inside |x|. Then can't we just write here:

-(x-3)+ sq. root of (-x*x)
=-x+3+ sq. root of (-(x)^2)
= -x+3+(-(x)^2)^1/2
=-x+3+|-1|*x
= -x+3+x
=3

When I am removing | |, I am just considering positive value of whatever is inside | |.

May I ask a very basic question here?

We sat that absolute value of any number is positive. So shouldn't we write only positive value after removing the bars.
i.e. |x|= x

It's a very basic question but because of this only I am making mistakes in absolute value questions.

_________________
Director  G
Joined: 02 Sep 2016
Posts: 625
Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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So what I understood is:

Absolute value of x is positive x when x is >=0.
Absolute value of x is negative x when x<0.

So we can't just say that absolute value of every number is positive no matter what is inside. The sign inside || will matter when || are removed.

|x| when x is positive= x
|x| when x is negative= -1*-x= +x
Math Expert V
Joined: 02 Sep 2009
Posts: 64073
Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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Shiv2016 wrote:
So what I understood is:

Absolute value of x is positive x when x is >=0.
Absolute value of x is negative x when x<0.

So we can't just say that absolute value of every number is positive no matter what is inside. The sign inside || will matter when || are removed.

|x| when x is positive= x
|x| when x is negative= -1*-x= +x

Your understanding is wrong. The post here: https://gmatclub.com/forum/if-x-0-what- ... l#p1847895 explains it.

Absolute value of a number CANNOT be negative! |x| is ALWAYS more than or equal to zero.

|some positive value| = that positive value

|some negative value| = -(that negative value), which still will be positive because -negative = positive.

|0| = 0.

Hope it's clear.
_________________
Director  G
Joined: 02 Sep 2016
Posts: 625
Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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BrushMyQuant wrote:
Whenever we have absolute value question and the variable is given as negative (like x<0 in this case) then the easiest way to solve the problem is to
take x=-k (where k is positive), this will make opening absolute values easy

[(x-3)^4]^1/4 + (-x|x|)^1/2
= [((-k)-3)^4]^1/4 + (-(-k)|-k|)^1/2
= [(-(k+3))^4]^1/4 + ((k)|-k|)^1/2
= [(k+3)^4]^1/4 + ((k)(k))^1/2 [as k is positive so |-k| will be equal to k]
= (k+3) + (k^2)^1/2
= k+3 + k = 2k + 3
[x = -k => k = -x]
= 2(-x) + 3 = 3-2x
Hope it helps!

siddharthsinha123 wrote:
If x<0

What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2

A. -3
B. x+3
C. 3-2x
D. 2x-3
E. 2

Can't we just cross off power 4 and 1/4?

x=-k

[(-k-3)^4]^1/4 + \sqrt{-(-k)*|-k|}
= -k-3+ k
= -3

I have tried this question so many times but still not really able to understand the minor things. I don't understand where I am going wrong: in cancelling the powers or opening the ||.
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Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  [#permalink]

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_________________ Re: If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3   [#permalink] 31 Mar 2020, 13:25

# If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3  