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Re: \sqrt (x+y)(x-y) [#permalink]
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Carcass wrote:
I am , instead, a big fun of you :)

Yeah, me too :) :wink:
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Re: \sqrt (x+y)(x-y) [#permalink]
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Thanks!!!! :banana
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Re: \sqrt (x+y)(x-y) [#permalink]
Ans: C

Cant we solve the RHS (QA) and we get (x-y)....
QA : (x-y)
QB : (x-y)

will the terms not cancel out on both sides irrespective of the signs ?
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Re: \sqrt (x+y)(x-y) [#permalink]
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anuppatel012 wrote:
Ans: C

Cant we solve the RHS (QA) and we get (x-y)....
QA : (x-y)
QB : (x-y)

will the terms not cancel out on both sides irrespective of the signs ?


My point about this question needing some information that ensures x and y are non-negative is best shown with this example:
Notice that \((\sqrt{x})^2=x\) in MANY cases.
For example, if x = 4, we get: \((\sqrt{4})^2=2^2=4\)
And, if x = 25, we get: \((\sqrt{25})^2=5^2=25\)

However, the same cannot be said if x is a NEGATIVE number.
For example, if x = -1, we get: \((\sqrt{-1})^2=?^2\)
First off, we don't know the value of \(\sqrt{-1}\)
Second, we certainly can't say that \((\sqrt{-1})^2=-1\), since that would mean that it's possible for some number SQUARED to equal -1.

Does that help?

Cheers,
Brent
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Re: \sqrt (x+y)(x-y) [#permalink]
You cannot square root a negative # so that why the answer cannot be D.
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Re: \sqrt (x+y)(x-y) [#permalink]
brajones wrote:
You cannot square root a negative # so that why the answer cannot be D.


Great question!
The problem is that, when x is negative, √x is not a real number. So, we can't really compare the quantities.

For example, if we have...
QUANTITY A: √(-3)
QUANTITY B: 5
.....we can't compare the two quantities, since Quantity A is not an actual real number.
Comparing those two values would be analogous to ...
QUANTITY A: elephant (also not a number)
QUANTITY B: 5
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