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If x < 0, then root({-x} •|x|) is

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Re: modes and square root  [#permalink]

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New post 11 Feb 2011, 00:42
1
the question is: If x < 0, what is the value for \(sqrt{-x*\mid x \mid}\):

\(sqrt{(+x)*(+x)}\)
\(sqrt{x^2}\)
\(\pm x\)

Reasoning is:

Since x < 0; only -x is the valid value. +x can be ignored.

Ans: "A"
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Re: Square root and Mod Problem  [#permalink]

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New post 07 Mar 2011, 22:32
sqrt(x^2) = |x|
if x <0 |x| = -x
x^2 = -x|x|

Hence A
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Re: Square root and Modulus  [#permalink]

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New post 14 Jun 2011, 01:07
A as root will always give positive value.
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Re: If x < 0, then root({-x} •|x|) is  [#permalink]

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New post 29 Nov 2012, 13:10
x is < 0.

This makes sense, if it was positive, then it would be the square root of (-)(+)(|+|)= -. We can't have the square of a negative, not a real #

so x is -. Let's pick x= -5. (-1)(-5)(|-5|)= √25 = 5. answer is 5 which is -x.
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Re: Square root and Modulus  [#permalink]

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New post 29 Nov 2012, 13:51
Bunuel wrote:
udaymathapati wrote:
If x < 0, then \sqrt{-x} •|x|) is
A. -x
B. -1
C. 1
D. x
E. \sqrt{x}


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.



Sorry Bunuel but the approach :\(\sqrt{-x* |x|}\) if we put this one ^2 then we have simply \(-x * |x|\). The latter is positive, so we have a quantity straight negative \(- X\).

or is wrong this simple way ))

thanks
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Re: modes and square root  [#permalink]

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New post 29 Nov 2012, 13:55
fluke wrote:
the question is: If x < 0, what is the value for \(sqrt{-x*\mid x \mid}\):

\(sqrt{(+x)*(+x)}\)
\(sqrt{x^2}\)
\(\pm x\)

Reasoning is:

Since x < 0; only -x is the valid value. +x can be ignored.

Ans: "A"



Sorry fluke but the quantity \(\sqrt{x^2}\) is not + or - X but only + X, if we have \(x^2= something\) then we have \(+ or -\).........
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Re: If x < 0, then root({-x} •|x|) is  [#permalink]

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New post 03 Dec 2012, 19:59
Since \(x < 0\),
\(\sqrt{-x|x|}\)
\(\sqrt{x^2}=|x|\)

Answer: -x since x is of negative value.
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Re: If x < 0, then root({-x} •|x|) is  [#permalink]

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New post 08 May 2013, 20:43
I said X before I read the very important point that x<-1

I need to stop that
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Re: Square root and Modulus  [#permalink]

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New post 15 Jun 2013, 08:19
1
How can the solution be negative if we're taking the square root of a positive number?

Bunuel wrote:
mbafall2011 wrote:
udaymathapati wrote:
If x < 0, then \sqrt{-x} •|x|) is
A. -x
B. -1
C. 1
D. x
E. \sqrt{x}


what is the source of this question. I havent seen any gmat question testing imaginary numbers


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.
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Re: Square root and Modulus  [#permalink]

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New post 15 Jun 2013, 08:23
WholeLottaLove wrote:
How can the solution be negative if we're taking the square root of a positive number?


Please read the solution carefully: if-x-0-then-root-x-x-is-100303.html#p773754

The answer is A, which is \(-x\), since \(x\) is negative then \(-x=-(negative)=positive\).
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Re: Square root and Modulus  [#permalink]

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New post 15 Jun 2013, 08:25
Ha! I got it just before I read your response. That is a very tricky problem - it's -(-x). Thanks!

Bunuel wrote:
WholeLottaLove wrote:
How can the solution be negative if we're taking the square root of a positive number?


Please read the solution carefully: if-x-0-then-root-x-x-is-100303.html#p773754

The answer is A, which is \(-x\), since \(x\) is negative then \(-x=-(negative)=positive\).
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If x < 0, then root({-x} •|x|) is  [#permalink]

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New post 03 Apr 2017, 09:42
Bunuel wrote:
Square root function can not give negative result


Bunuel, I do not understand this. When you graph \(x^2 - 9 = 0\) you will get a parabola with roots at 3 and -3. That is because \(\sqrt{9}\) is 3 or -3, not just 3. When you square root a number, the answer will always be plus/minus because when you square a negative number, it becomes positive. For example, \(\sqrt{25}\) is 5 or -5, not just 5. This is because \(5^2\) and \((-5)^2\) both equal 25. Why is this different for this question?

I think what fluke wrote is correct:

fluke wrote:
the question is: If x < 0, what is the value for \(sqrt{-x*\mid x \mid}\):

\(sqrt{(+x)*(+x)}\)
\(sqrt{x^2}\)
\(\pm x\)

Reasoning is:

Since x < 0; only -x is the valid value. +x can be ignored.

Ans: "A"
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Re: If x < 0, then root({-x} •|x|) is  [#permalink]

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New post 03 Apr 2017, 09:49
joondez wrote:
Bunuel wrote:
Square root function can not give negative result


Bunuel, I do not understand this. When you graph \(x^2 - 9 = 0\) you will get a parabola with roots at 3 and -3. That is because \(\sqrt{9}\) is 3 or -3, not just 3. When you square root a number, the answer will always be plus/minus because when you square a negative number, it becomes positive. For example, \(\sqrt{25}\) is 5 or -5, not just 5. This is because \(5^2\) and \((-5)^2\) both equal 25. Why is this different for this question?

I think what fluke wrote is correct:

fluke wrote:
the question is: If x < 0, what is the value for \(sqrt{-x*\mid x \mid}\):

\(sqrt{(+x)*(+x)}\)
\(sqrt{x^2}\)
\(\pm x\)

Reasoning is:

Since x < 0; only -x is the valid value. +x can be ignored.

Ans: "A"


What I said there is a fact.

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root. That is, \(\sqrt{16}=4\), NOT +4 or -4. Even roots have only a positive value on the GMAT.

In contrast, the equation \(x^2=16\) has TWO solutions, +4 and -4.

Odd roots have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).
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Re: If x < 0, then root({-x} •|x|) is  [#permalink]

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New post 05 Aug 2018, 12:50
Bunuel wrote:
udaymathapati wrote:
If x < 0, then \sqrt{-x} •|x|) is
A. -x
B. -1
C. 1
D. x
E. \sqrt{x}


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.



I have a different theory here. They explicitly state that x is negative. Then they give us an eq. Your solution neglects the fact that x is negative.
\(\sqrt{-x*|x|}=\sqrt{-(negative)*|negative|)}=\sqrt{positive*positive}=|x|=|x|\)
There are two solutions for |x| and both of them are in the answers. Some thing is off with this question imho
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Re: If x < 0, then root({-x} •|x|) is  [#permalink]

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New post 05 Aug 2018, 21:27
nobilisrex wrote:
Bunuel wrote:
udaymathapati wrote:
If x < 0, then \sqrt{-x} •|x|) is
A. -x
B. -1
C. 1
D. x
E. \sqrt{x}


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.



I have a different theory here. They explicitly state that x is negative. Then they give us an eq. Your solution neglects the fact that x is negative.
\(\sqrt{-x*|x|}=\sqrt{-(negative)*|negative|)}=\sqrt{positive*positive}=|x|=|x|\)
There are two solutions for |x| and both of them are in the answers. Some thing is off with this question imho


To check your theory, I'd suggest to plug some negative number and check what you'd get.

P.S. The question as well as the solution is fine.
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Re: If x < 0, then root({-x} •|x|) is  [#permalink]

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New post 19 Sep 2018, 17:38
Bunuel wrote:
udaymathapati wrote:
If x < 0, then \sqrt{-x} •|x|) is
A. -x
B. -1
C. 1
D. x
E. \sqrt{x}


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.


since we are given x<0, this means x is -ve. therefore:
1. -x--> -(-x) --> x

so,
2. sqrt(-x . |x|) --> sqrt(x . |x|)

can you explain what is wrong with stmt 1?
i lose the plot on stmt 2....can you explain why |x|--> -x? it should be +x

thanks
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If x < 0, then root({-x} •|x|) is  [#permalink]

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New post 20 Sep 2018, 08:35
Bunuel wrote:
udaymathapati wrote:
If x < 0, then \sqrt{-x} •|x|) is
A. -x
B. -1
C. 1
D. x
E. \sqrt{x}


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.



I understood the concept, However I don't understand how you did this \(\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}\)
Inside mod X can be either positive or negative, how can you assume it to be negative?
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