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GMAT Tutor S
Joined: 24 Jun 2008
Posts: 1526
Re: No. Properties  [#permalink]

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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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SVP  Joined: 07 Nov 2007
Posts: 1614
Location: New York
Re: No. Properties  [#permalink]

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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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Smiling wins more friends than frowning
Manager  Joined: 04 Jan 2009
Posts: 211
Re: No. Properties  [#permalink]

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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
Manager  Joined: 05 Oct 2008
Posts: 241
Re: No. Properties  [#permalink]

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Thanks Ian for the explanation and for posting the question in mathematical format.

+2
Senior Manager  Joined: 20 Mar 2008
Posts: 414
Root of the product of -x and |x|  [#permalink]

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Root of the product of -x and |x|
Attachments root x mod x.JPG [ 9.55 KiB | Viewed 5473 times ]

Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: Root of the product of -x and |x|  [#permalink]

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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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Senior Manager  Joined: 20 Mar 2008
Posts: 414
Re: Root of the product of -x and |x|  [#permalink]

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+1 to you.

Thanks for the explanation. When I saw this problem, I had a feeling there was something I was missing.
Intern  Joined: 22 Sep 2009
Posts: 34
Re: Root of the product of -x and |x|  [#permalink]

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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

I don't follow why y be positive? Bunuel please explain
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: Root of the product of -x and |x|  [#permalink]

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2
2
pradhan wrote:
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

I don't follow why y be positive? Bunuel please explain

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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Intern  Joined: 28 Apr 2009
Posts: 30
Re: Root of the product of -x and |x|  [#permalink]

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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 ?
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: Root of the product of -x and |x|  [#permalink]

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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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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
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: Square root - positive or negative?  [#permalink]

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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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Senior Manager  Joined: 22 Dec 2009
Posts: 291
Re: Square root - positive or negative?  [#permalink]

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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 ---- 

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

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 $$\sqrt{4} = 2$$ ---- 

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

Hence if we subsitute the values of $$\sqrt{4}$$ from equation 2 ( i.e. $$\sqrt{-x*|x|}$$ ) and value of 2 from equation  ( i.e. -x )... in equation , 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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Re: if x<0  [#permalink]

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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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