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Intern  Joined: 08 Jul 2009
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If x<0, then (-x*|x|)^(1/2) is:  [#permalink]

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100 00:00

Difficulty:   45% (medium)

Question Stats: 51% (00:47) correct 49% (00:47) wrong based on 1641 sessions

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If x < 0, then $$\sqrt{-x|x|}$$ is:

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

Originally posted by nss123 on 29 Jul 2009, 16:42.
Last edited by Bunuel on 28 Aug 2019, 23:19, edited 1 time in total.
Renamed the topic and edited the question.
Math Expert V
Joined: 02 Sep 2009
Posts: 59586
If x<0, then (-x*|x|)^(1/2) is:  [#permalink]

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29
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saxenashobhit wrote:

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

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 $$\sqrt{x}$$, 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, $$\sqrt{125} =5$$ and $$\sqrt{-64} =-4$$.

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 $$x=-5<0$$ --> $$\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x$$.

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

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9
The easiest way for you to solve this problem would be to plug in a number and see what happens.

Let's say $$x = -1$$

$$\sqrt{-x|x|}=\sqrt{-(-1)|-1|}=\sqrt{(1)(1)}=1 = -(-1)$$

##### General Discussion
Intern  Joined: 02 Aug 2009
Posts: 1
Re: If x<0, then (-x*|x|)^(1/2) is:  [#permalink]

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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.
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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 $$|x|=-x$$ --> $$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$.

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

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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 $$\sqrt{x}$$, 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, $$\sqrt{125} =5$$ and $$\sqrt{-64} =-4$$.

Solution for the original question:

Given: $$x<0$$ Question: $$\sqrt{-x*|x|}=?$$.

Remember: $$\sqrt{x^2}=|x|$$.

As $$x<0$$, then $$|x|=-x$$ --> $$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$.

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

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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 $$x=-5<0$$ --> $$\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x$$.

Hope it's clear.
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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.

$$\sqrt{4}$$ = + 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
$$\sqrt{-x*|x}|$$ . 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, $$\sqrt{125} =5$$ and $$\sqrt{-64} =-4$$.

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

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

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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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 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

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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

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

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_________________ Re: If x<0, then (-x*|x|)^(1/2) is:   [#permalink] 28 Aug 2019, 23:02
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