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square roots and absolute values

Expert replies
Source: — Problem Solving |

by liferocks » Mon May 10, 2010 7:15 pm
sqrt(-x|x|)=sqrt(-x*(-x))..since x<0
=sqrt(x^2)
=|x|
=-x

ans option A
"If you don't know where you are going, any road will get you there."
Lewis Carroll
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by aturpening » Mon May 10, 2010 7:58 pm
It was my understanding that |x| was the distance along a number line that x was from zero, and since x cannot be a negative distance from zero, |x| had to be a positive number no matter whether x was negative or not? How can |x| be equal to -x?
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by aturpening » Mon May 10, 2010 8:05 pm
By the way, your answer is correct.
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by sumanr84 » Mon May 10, 2010 8:38 pm
This is absurd as |x| is always > 0 ( whether X is <0 or >0)
I am on a break !!
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by liferocks » Mon May 10, 2010 8:52 pm
aturpening wrote:It was my understanding that |x| was the distance along a number line that x was from zero, and since x cannot be a negative distance from zero, |x| had to be a positive number no matter whether x was negative or not? How can |x| be equal to -x?
your understanding is absolutely correct.Now if we use that logic to get value of |x| we will get

|x|=x..when x>0...ex 1,2,3 etc
=0 when x=0
=-x ..when x<0 ..because since sign of x is -ve..if we multiply it with -1 it will give a positive value..ex -1,-2 etc which will make |x| positive

hope this clarifies the confusion
"If you don't know where you are going, any road will get you there."
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by gmatmachoman » Tue May 11, 2010 1:26 am
aturpening wrote:It was my understanding that |x| was the distance along a number line that x was from zero, and since x cannot be a negative distance from zero, |x| had to be a positive number no matter whether x was negative or not? How can |x| be equal to -x?
I would suggest picking numbers.

Try X = -2 ( as X<0)

|X| = 2

-X(|X|

-(-2)(2)

= 4

Sqrt (-X(|X|)

4^1/2

2 which is nothing but -X

Pick A
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by kstv » Tue May 11, 2010 2:15 am
aturpening wrote:If x<0, then sqrt(-x|x| ) is
(A) -x (B) -1 (C) 1 (D) x (E) sqrt(x)
x < 0 , |x| = x
-x = -(-x) = x

√-x |x|
= √x*x = √x² = +- x

since x < 0 option (A)
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by sanju09 » Tue May 11, 2010 2:22 am
sumanr84 wrote:This is absurd as |x| is always > 0 ( whether X is <0 or >0)
If x < 0, -x > 0, let's call -x some positive number y, now |x| is also y and y × y = y^2, whose square root is positive [spoiler]y or -x[/spoiler].

[spoiler]A[/spoiler]
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