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Is Sqrt (x-5)^2 = 5 - x

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Intern
Joined: 04 Sep 2005
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Is Sqrt (x-5)^2 = 5 - x [#permalink]  22 Sep 2005, 12:41
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Is Sqrt (x-5)^2 = 5 - x ?

1) -x |x| > 0

2 ) 5 - x > 0

OG explanation starts by saying that Sqrt (y)^2 = |y|. I can't understand that statement. Isn't Sqrt (y)^2 = + or - y ???

I feel i'm missing something here...any help would be appreciated.
Manager
Joined: 06 Aug 2005
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[#permalink]  22 Sep 2005, 13:15
No sqrt(x^2) = x
and sqrt ((-x)^2) = x

Exactly how was the expression on the left formatted ?
VP
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Re: OG DS Question [#permalink]  22 Sep 2005, 14:38
zag1979 wrote:
I have a problem understanding the solution of this DS problem:

Is Sqrt (x-5)^2 = 5 - x ?

1) -x |x| > 0

2 ) 5 - x > 0

OG explanation starts by saying that Sqrt (y)^2 = |y|. I can't understand that statement. Isn't Sqrt (y)^2 = + or - y ???

I feel i'm missing something here...any help would be appreciated.

is it D.

well to me, what they are saying is:
sqrt(y * y) = y

in this case we omit -y as possible result since we know beforehand number being squared. in other words:
sqrt(4) = +2/-2 but
sqrt(-2 * -2) = -2 or
sqrt(2 * 2) = 2
Director
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[#permalink]  22 Sep 2005, 18:11
I can see that sqrt (x-5)*(x-5) is x-5. But, as duttsit, I think the answer is D.
Intern
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OG DS Question [#permalink]  22 Sep 2005, 23:33
OA is D.

The expression on the left is formatted this way:

Sqrt ((x-5)^2)

What the OE basically says is that Sqrt ((x-5)^2) = x - 5 if x - 5 is positive and it is 5 - x if x - 5 is negative.

If I go by the same logic then Sqrt ((-2).(-2)) = 2 and cannot be equal to -2. I don't really see the reason behind that. Why couldn't the root be -2?
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Re: OG DS Question [#permalink]  23 Sep 2005, 00:12
zag1979 wrote:
If I go by the same logic then Sqrt ((-2).(-2)) = 2 and cannot be equal to -2. I don't really see the reason behind that. Why couldn't the root be -2?

The official rule is that the sqrt(x^2) = abs(x) for every real number x.

Negatives inside sqrts mean imaginary numbers. Not on the test.

Edit: Hopefully I haven't added to the confusion - it's late & my brain is fried. If you're lost, check out basic imaginary number stuff. IE, sqrt(-1) = i. ... Again, NOT on the test.
Re: OG DS Question   [#permalink] 23 Sep 2005, 00:12
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