If x

Question

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

BrushMyQuant · Accepted Answer

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

https://www.youtube.com/watch?v=APc3d9-f__Y

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

\sqrt {(x-3)^4} + \sqrt{-x|x|} 
=  \sqrt {(-k-3)^4} + \sqrt{-(-k)|-k|} 
=  \sqrt {-(k+3)^4} + \sqrt{+ k|k|}  ( as |-k| = |k|  )
=  \sqrt {(k+3)^4} + \sqrt{+k*k}  ( as |k| = k, because k>0  )
=  \sqrt {(k+3)^4} + \sqrt{k^2} 
= k + 3 + k
= 2k + 3
 
=> 2k + 3= 2(-x) + 3 = 3-2x

So,  Answer will be C 
Hope it helps!

Watch the following video to MASTER Absolute Values

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

chetan2u · Answer

Hi

Best way is to substitute a value for X..
Let x be -2..
So  ^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

Bunuel · Answer

Formatted the question.

abhishekmayank · Answer

Hi,

Let me understand that why  ^1/4 = 5. I think that  ^1/4 should be -5

umg · Answer

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

khushbumodi · Answer

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

Sent from my SM-N9200 using  GMAT Club Forum mobile app

umg · Answer

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) .

Bunuel · Answer

Two things to remember:

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

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

Answer: C.

vmelgargalan · Answer

I am confused. How can you substitute x for a negative number, when x<0? wouldn't x have to be positive?

Shiv2016 · Answer

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.

Shiv2016 · Answer

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 | |.

Please help.

BrushMyQuant · Answer

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)
So final answer is
-x + 3 - x = 3-2x

Hope it helps!

Shiv2016 · Answer

Thank you so much for your reply.
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.

BrushMyQuant · Answer

|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 · Answer

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

Bunuel · Answer

Shiv2016 · Answer

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

x=-k

^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 ||.

Archit3110 · Answer

given x<0
so let x= -1
we get  \sqrt {(x-3)^4} + \sqrt{-x|x|}  
this will be 5
substitute value of x =-1 in answer options
option C is correct

bumpbot · Answer

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

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