If x < 0, then (-x*|x|)^(1/2) is:

SOME NOTES:

1. GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.

2. Any nonnegative real number has a unique non-negative square root called the principal square root and unless otherwise specified, the square root is generally taken to mean the principal square root.

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 non-negative value on the GMAT.

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

3. \(\sqrt{x^2}=|x|\).

The point here is that as square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\).

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: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).


BACK TO THE ORIGINAL QUESTION:

1. If \(x<0\), then \(\sqrt{-x*|x|}\) equals:
A. \(-x\)
B. \(-1\)
C. \(1\)
D. \(x\)
E. \(\sqrt{x}\)

\(\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x\). Note that as \(x<0\) then \(-x=-negative=positive\), so \(\sqrt{-x*|x|}=-x=positive\) as it should be.

Or just substitute the some negative \(x\), let \(x=-5<0\) --> \(\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x\).

Answer: A.

Hope it's clear.
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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.
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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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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.

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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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.

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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I had same question today

\(\sqrt{4}\) = + or - 2. So answer should be + or - x. I don't get how can OA be -x

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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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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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:

GMAT assassins aren't born, they're made,
Rich

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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Let's take x= -5
=>So √ (-x |x|) = √ (-(-5)|-5|)
=> √ (5*5) = √ (25)
= 5
= -(-5)
= -x
(option a)

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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Hi,
Isn't the output of mod supposed to be positive?

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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