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CTG1983
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Is Sqrt[(X-3)^2]=3-X ?

i)X !=3 (!= means NOT EQUAL TO)

ii) -XIXI >0 (I I means modulus)

Vote B

The stem can be restated as: Is (3-X) >= 0, since Sqrt[(X-3)^2] is the principal root of (X-3)^2 and is non-negative

 

I: Clearly insuff

 

II: -X*|X| > 0

=> X 0

=> 3-X > 0

Suff

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Vote B

The stem can be restated as: Is (3-X) >= 0, since Sqrt[(X-3)^2] is the principal root of (X-3)^2 and is non-negative

 

I: Clearly insuff

 

II: -X*|X| > 0

=> X 0

=> 3-X > 0

Suff

 

arjmen, can u plz make the bold part a bit clearer? Wht's principal root?

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  • 4 weeks later...

Hi arjmen/Bob

 

I couldn't make out on how exactly you came to equation Is (3-X) >= 0 ? Can you plz explain it a bit more.

 

From the question I should check that Is IX-3I = 3-X ? Using this equation I was able to find the answer, but I'm interseted in knowing your way.

 

Thanks a lot

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I couldn't make out on how exactly you came to equation Is (3-X) >= 0 ? Can you plz explain it a bit more.
Take a positive number, square it, and then take the principal square root of the result... you will end up with the number you started with. Example:

 

Take 5.

Square it: 5^2 = 25.

Take the principal square root of the result: sqrt(25) = 5.

You end up with the number you started with.

 

On the other hand, start with a negative number, square it, and then take the principal square root of the result... you will end up with the opposite of the number you started with. Example:

 

Take -5.

Square it: (-5) = 25.

Take the principal square root of the result: sqrt(25) = 5.

You end up with the opposite of the number you started with.

 

So, to generalize:

sqrt(x^2) = x if x > 0

sqrt(x^2) = -x if x

 

To put it even more simply:

sqrt(x^2) = |x|.

 

So, turning to the question at hand...

sqrt(x-3) = x-3 if x-3 > 0

sqrt(x-3) = -(x-3) = 3-x if x-3

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