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Re: If x < 0, then (-x*|x|)^(1/2) is: [#permalink]
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LM wrote:
If x<0, then \(\sqrt{-x|x|}\) is:

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


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.
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Re: If x < 0, then (-x*|x|)^(1/2) is: [#permalink]
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.
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Re: If x < 0, then (-x*|x|)^(1/2) is: [#permalink]
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.
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Re: If x < 0, then (-x*|x|)^(1/2) is: [#permalink]
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shammokando wrote:
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.


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 \(\sqrt[4]{x}\), 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, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).

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.
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Re: If x < 0, then (-x*|x|)^(1/2) is: [#permalink]
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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.
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Re: If x < 0, then (-x*|x|)^(1/2) is: [#permalink]
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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: If x < 0, then (-x*|x|)^(1/2) is: [#permalink]
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hailtothethief23 wrote:
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.

I had same question today

\(\sqrt{4}\) = + or - 2. So answer should be + or - x. I don't get how can OA be -x
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Re: If x < 0, then (-x*|x|)^(1/2) is: [#permalink]
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shrive555 wrote:
Bunel: i've already seen all the explanation just one more question.

as If \(x<0\), then \(\sqrt{-x*|x|}\)
Ans is -x
is Answer of the question depends on the condition x<0 or it depends on the sqrt (even root)

lets keep the condition same i.e x<0 and take odd root say cube root. i.e
\(\sqrt[3]{-x*|x}|\) . what would be the answer, would it be x then ?

Thanks


About the odd roots: odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).

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 \(\sqrt[3]{-x*|x}|\) will equal neither to x nor to -x: \(\sqrt[3]{-x*|x|}=\sqrt[3]{x^2}=x^{\frac{2}{3}}\), for example if \(x=-8<0\) then \(\sqrt[3]{-x*|x|}=\sqrt[3]{x^2}=\sqrt[3]{64}=4\).

If it were \(x<0\) and \(\sqrt[3]{-x^2*|x}|=?\), then \(\sqrt[3]{-x^2*|x}|=\sqrt[3]{-x^2*(-x)}=\sqrt[3]{x^3}=x<0\). Or substitute the value let \(x=-5<0\) --> \(\sqrt[3]{-x^2*|x}|=\sqrt[3]{-25*5}=\sqrt[3]{-125}=-5=x<0\).

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.
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Re: If x < 0, then (-x*|x|)^(1/2) is: [#permalink]
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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.
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Re: If x < 0, then (-x*|x|)^(1/2) is: [#permalink]
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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:

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Re: If x < 0, then (-x*|x|)^(1/2) is: [#permalink]
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nss123 wrote:
If x<0, then \(\sqrt{-x|x|}\) is:

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


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
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Re: If x < 0, then (-x*|x|)^(1/2) is: [#permalink]
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Let's take x= -5
=>So √ (-x |x|) = √ (-(-5)|-5|)
=> √ (5*5) = √ (25)
= 5
= -(-5)
= -x
(option a)
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Re: If x < 0, then (-x*|x|)^(1/2) is: [#permalink]
Hi brunel, Shouldn't −(−5) become +5?
Thanks for great answer always and your time in advanced.
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Re: If x < 0, then (-x*|x|)^(1/2) is: [#permalink]
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Kimberly77 wrote:
Hi brunel, Shouldn't −(−5) become +5?
Thanks for great answer always and your time in advanced.


Yes, -(-5) = 5, which is -x (x = -5).
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Re: If x < 0, then (-x*|x|)^(1/2) is: [#permalink]
Hi,
Isn't the output of mod supposed to be positive?
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Re: If x < 0, then (-x*|x|)^(1/2) is: [#permalink]
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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

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