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If x <0 , then \sqrt{-x*|x|} equals 1. -x 2. -1 3. 1 4.

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25 Jan 2009, 05:40
1
If $$x < 0$$

$$\sqrt{-x |x|} =$$

Notice that $$x$$ is negative, so the answer cannot be $$x$$ here. We are finding the positive square root of something in this question, so the answer cannot be negative.

If we simplify what is underneath the root, we have $$\sqrt{ x^2}$$, which is equal to $$x$$ if $$x$$ is positive . If, on the other hand, $$x$$ is negative, then $$\sqrt{x^2}$$ is equal to $$- x$$.
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25 Jan 2009, 07:50
study wrote:
If x<0, then [sqrt (-x absolute x )] is
(please note that both variables - minus x and absolute x - come under the root sign)

1
-1
sqrt x
x
-x

let say -x=y (y is positive)

sqrt (-x |x|) = sqrt (y|-y|) = sqrt (y^2) = y or -y

when y +ve --> answer =y = -x
wnen y -ve --> answer = -y=x

here y is +ve so answer =-x

E
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25 Jan 2009, 09:44
yes sir. sqrt(f) has to be positive otherwise, we run into the following problem:
2 is sqrt(4)
if we say sqrt(4)=-2 or 2
in one situation we run into situation where 2+2=0.
So, unless we are asked to find (-sqrt(f)) sqrt(f) always yields positive number.

x2suresh wrote:
study wrote:
If x<0, then [sqrt (-x absolute x )] is
(please note that both variables - minus x and absolute x - come under the root sign)

1
-1
sqrt x
x
-x

let say -x=y (y is positive)

sqrt (-x |x|) = sqrt (y|-y|) = sqrt (y^2) = y or -y

when y +ve --> answer =y = -x
wnen y -ve --> answer = -y=x

here y is +ve so answer =-x

E

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tusharvk
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25 Jan 2009, 12:05
Thanks Ian for the explanation and for posting the question in mathematical format.

+2
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Root of the product of -x and |x|  [#permalink]

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31 Oct 2009, 22:40
Root of the product of -x and |x|
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Re: Root of the product of -x and |x|  [#permalink]

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31 Oct 2009, 22:52
3
Jivana wrote:
Root of the product of -x and |x|

$$x<0$$, $$y=\sqrt{-x*|x|}$$?

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

When we see the equation of a type: $$y=\sqrt{x^2}$$ then $$y=|x|$$, which means that $$y$$ can not be negative but $$x$$ can.
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Re: Root of the product of -x and |x|  [#permalink]

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31 Oct 2009, 22:59
+1 to you.

Thanks for the explanation. When I saw this problem, I had a feeling there was something I was missing.
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Re: Root of the product of -x and |x|  [#permalink]

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01 Nov 2009, 05:38
x<0 ....... let a=-x. so a is +ve

hence y= sqrt{-x*|x|}=sqrt{a*|-a|}
y = sqrt{a*a}
y = +a or -a
y=-x or x

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Re: Root of the product of -x and |x|  [#permalink]

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01 Nov 2009, 06:12
2
2
x<0 ....... let a=-x. so a is +ve

hence y= sqrt{-x*|x|}=sqrt{a*|-a|}
y = sqrt{a*a}
y = +a or -a
y=-x or x

Here is my solution:
Root of the product of -x and |x|[/quote]

$$x<0$$, $$y=\sqrt{-x*|x|}$$?

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

When we see the equation of a type: $$y=\sqrt{x^2}$$ then $$y=|x|$$, which means that $$y$$ can not be negative but $$x$$ can.

I say that $$y$$can not be negative as $$y=\sqrt{some expression}$$, and even root from the expression (some value) is never negative (as for GMAT we are dealing only with real numbers).

Consider this $$y=\sqrt{16}$$ --> $$y=4$$ and not $$y=4$$ or $$y=-4$$.

Different case when $$x^2=16$$, in this case $$x=4$$ or $$x=-4$$.

In original question we have $$y=\sqrt{x^2}$$, which means $$y=|x|$$ (you can see that $$y$$ is equal to the absolute value of $$x$$, and we know that absolute value is never negative). But we know that $$x<0$$, so $$y=|x|=-x$$. Look at $$x$$, since it's negative $$-x$$ would be positive and we have $$y=some positive value$$, which is true.

Hope it's clear now.
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Re: Root of the product of -x and |x|  [#permalink]

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14 Nov 2009, 07:11
Bunuel wrote:
I say that $$y$$can not be negative as $$y=\sqrt{some expression}$$, and even root from the expression (some value) is never negative (as for GMAT we are dealing only with real numbers).

Consider this $$y=\sqrt{16}$$ --> $$y=4$$ and not $$y=4$$ or $$y=-4$$.

But why ?? is this a standard formula ?
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Re: Root of the product of -x and |x|  [#permalink]

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14 Nov 2009, 07:24
mbaquestionmark wrote:
Bunuel wrote:
I say that $$y$$can not be negative as $$y=\sqrt{some expression}$$, and even root from the expression (some value) is never negative (as for GMAT we are dealing only with real numbers).

Consider this $$y=\sqrt{16}$$ --> $$y=4$$ and not $$y=4$$ or $$y=-4$$.

But why ?? is this a standard formula ?

When the GMAT provides the square root sign for an even root, such as a square root, then the only accepted answer is the positive root.

That is, $$\sqrt{16} = 4$$, NOT +4 or -4. In contrast, the equation x^2 = 16 has TWO solutions, +4 and -4.

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, CBRT(64) = 4, CBRT(-27) = -3.

Hope it's clear.
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Re: Square root - positive or negative?  [#permalink]

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31 Dec 2009, 04:22
I think the answer is 4. X

Coz: given that x<0 , -(-x) is positive and module x is also positive so we get x^2 , and end result is x

If you do not think so please correct me
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Re: Square root - positive or negative?  [#permalink]

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31 Dec 2009, 07:57
sher1978 wrote:
I think the answer is 4. X

Coz: given that x<0 , -(-x) is positive and module x is also positive so we get x^2 , and end result is x

If you do not think so please correct me

Answer to this question is A (-x). Please refer to the solutions above. Do ask if anything remains unclear.
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Re: Square root - positive or negative?  [#permalink]

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31 Dec 2009, 12:46
I guess the best way to solve this question is to plug in values and see how the result features.

Before we do the same, explanation by HongHu is perfect. (Cant post the link to his reply as I am new to the club )

i.e Square root of any positive number is a positive number

Now getting back to the actual question:

If x<0, then $$\sqrt{-x*|x|}$$ equals:
1. -x
2. -1
3. 1
4. x
5. $$\sqrt{x}$$

Approach:
Since x <0, let use plug in a negative integer as x, for e.g.

Let x = -2 ---- [1]

Hence $$\sqrt{-x*|x|}= \sqrt{-(-2)*|-2|} = \sqrt{2*2} = \sqrt{4}$$ ---- [2]

Now as per the discussion going on, the point of disagreement is that $$\sqrt{4}$$ = 2 or -2. But as stated by HongHu, we would consider $$\sqrt{4}$$ = 2 only! The reason for this conclusion is that the radical sign used in $$sqrt{x}$$ is used to show the positive square root of a number x.

Therefore $$\sqrt{4} = 2$$ ---- [3]

Now as per Equation [1] marked above, x = -2 which can also be written as -x = 2 ---- [4]

Hence if we subsitute the values of $$\sqrt{4}$$ from equation 2 ( i.e. $$\sqrt{-x*|x|}$$ ) and value of 2 from equation [4] ( i.e. -x )... in equation [3], we get the following result:

$$\sqrt{-x*|x|} = -x$$ and hence the correct answer is 1

This approach can be verified by another question which featured in the GMAT prep exams...

If $$x \neq 0$$, then $$\frac{\sqrt{x^2}}{x}$$ equals:
1. -1
2. 0
3. 1
4. x
5. $$\frac{|x|}{x}$$

Approach:
In this question, we don't know whether x is negative or positive. Hence $$\sqrt{x^2}$$ is always equal to |x|. Reason for the same is: x is a variable here and we dont know the exact value. The absolute value sign is needed when we are taking the square root of a square of a variable, which may be positive or negative

Therefore in this situation, the answer would be 5

Hope this helps!

Cheers!
JT
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08 Jan 2010, 04:17
No it's (A): -x

$$\sqrt{(-x)|x|}=\sqrt{(-x)(-x)}=\sqrt{x^2}=|x| = -x$$ (because x<0)
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Re: If x <0 , then \sqrt{-x*|x|} equals 1. -x 2. -1 3. 1 4.  [#permalink]

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Re: If x <0 , then \sqrt{-x*|x|} equals 1. -x 2. -1 3. 1 4.   [#permalink] 20 Aug 2018, 06:40

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