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If x < 0, then root({-x} •|x|) is

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Re: modes and square root [#permalink]

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11 Feb 2011, 01:42
1
KUDOS
the question is: If x < 0, what is the value for $$sqrt{-x*\mid x \mid}$$:

$$sqrt{(+x)*(+x)}$$
$$sqrt{x^2}$$
$$\pm x$$

Reasoning is:

Since x < 0; only -x is the valid value. +x can be ignored.

Ans: "A"
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Re: Square root and Mod Problem [#permalink]

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07 Mar 2011, 23:32
sqrt(x^2) = |x|
if x <0 |x| = -x
x^2 = -x|x|

Hence A
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Re: Square root and Modulus [#permalink]

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14 Jun 2011, 02:07
A as root will always give positive value.
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Re: If x < 0, then root({-x} •|x|) is [#permalink]

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29 Nov 2012, 14:10
x is < 0.

This makes sense, if it was positive, then it would be the square root of (-)(+)(|+|)= -. We can't have the square of a negative, not a real #

so x is -. Let's pick x= -5. (-1)(-5)(|-5|)= √25 = 5. answer is 5 which is -x.
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Re: Square root and Modulus [#permalink]

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29 Nov 2012, 14:51
Bunuel wrote:
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.

Sorry Bunuel but the approach :$$\sqrt{-x* |x|}$$ if we put this one ^2 then we have simply $$-x * |x|$$. The latter is positive, so we have a quantity straight negative $$- X$$.

or is wrong this simple way ))

thanks
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Re: modes and square root [#permalink]

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29 Nov 2012, 14:55
fluke wrote:
the question is: If x < 0, what is the value for $$sqrt{-x*\mid x \mid}$$:

$$sqrt{(+x)*(+x)}$$
$$sqrt{x^2}$$
$$\pm x$$

Reasoning is:

Since x < 0; only -x is the valid value. +x can be ignored.

Ans: "A"

Sorry fluke but the quantity $$\sqrt{x^2}$$ is not + or - X but only + X, if we have $$x^2= something$$ then we have $$+ or -$$.........
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Re: If x < 0, then root({-x} •|x|) is [#permalink]

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03 Dec 2012, 20:59
Since $$x < 0$$,
$$\sqrt{-x|x|}$$
$$\sqrt{x^2}=|x|$$

Answer: -x since x is of negative value.
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Re: If x < 0, then root({-x} •|x|) is [#permalink]

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08 May 2013, 21:43
I said X before I read the very important point that x<-1

I need to stop that
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Re: Square root and Modulus [#permalink]

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15 Jun 2013, 09:19
How can the solution be negative if we're taking the square root of a positive number?

Bunuel wrote:
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: Square root and Modulus [#permalink]

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15 Jun 2013, 09:23
WholeLottaLove wrote:
How can the solution be negative if we're taking the square root of a positive number?

The answer is A, which is $$-x$$, since $$x$$ is negative then $$-x=-(negative)=positive$$.
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Re: Square root and Modulus [#permalink]

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15 Jun 2013, 09:25
Ha! I got it just before I read your response. That is a very tricky problem - it's -(-x). Thanks!

Bunuel wrote:
WholeLottaLove wrote:
How can the solution be negative if we're taking the square root of a positive number?

The answer is A, which is $$-x$$, since $$x$$ is negative then $$-x=-(negative)=positive$$.
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Re: If x < 0, then root({-x} •|x|) is [#permalink]

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25 Sep 2014, 09:07
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Re: If x < 0, then root({-x} •|x|) is [#permalink]

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29 Sep 2015, 10:45
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Re: If x < 0, then root({-x} •|x|) is [#permalink]

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10 Nov 2016, 06:44
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If x < 0, then root({-x} •|x|) is [#permalink]

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03 Apr 2017, 10:42
Bunuel wrote:
Square root function can not give negative result

Bunuel, I do not understand this. When you graph $$x^2 - 9 = 0$$ you will get a parabola with roots at 3 and -3. That is because $$\sqrt{9}$$ is 3 or -3, not just 3. When you square root a number, the answer will always be plus/minus because when you square a negative number, it becomes positive. For example, $$\sqrt{25}$$ is 5 or -5, not just 5. This is because $$5^2$$ and $$(-5)^2$$ both equal 25. Why is this different for this question?

I think what fluke wrote is correct:

fluke wrote:
the question is: If x < 0, what is the value for $$sqrt{-x*\mid x \mid}$$:

$$sqrt{(+x)*(+x)}$$
$$sqrt{x^2}$$
$$\pm x$$

Reasoning is:

Since x < 0; only -x is the valid value. +x can be ignored.

Ans: "A"
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Re: If x < 0, then root({-x} •|x|) is [#permalink]

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03 Apr 2017, 10:49
joondez wrote:
Bunuel wrote:
Square root function can not give negative result

Bunuel, I do not understand this. When you graph $$x^2 - 9 = 0$$ you will get a parabola with roots at 3 and -3. That is because $$\sqrt{9}$$ is 3 or -3, not just 3. When you square root a number, the answer will always be plus/minus because when you square a negative number, it becomes positive. For example, $$\sqrt{25}$$ is 5 or -5, not just 5. This is because $$5^2$$ and $$(-5)^2$$ both equal 25. Why is this different for this question?

I think what fluke wrote is correct:

fluke wrote:
the question is: If x < 0, what is the value for $$sqrt{-x*\mid x \mid}$$:

$$sqrt{(+x)*(+x)}$$
$$sqrt{x^2}$$
$$\pm x$$

Reasoning is:

Since x < 0; only -x is the valid value. +x can be ignored.

Ans: "A"

What I said there is a fact.

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, $$\sqrt{16}=4$$, NOT +4 or -4. Even roots have only a positive value on the GMAT.

In contrast, the equation $$x^2=16$$ has TWO solutions, +4 and -4.

Odd roots have the same sign as the base of the root. For example, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.
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Re: If x < 0, then root({-x} •|x|) is   [#permalink] 03 Apr 2017, 10:49

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