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I got the roots for the equation as (x-3)^2 (reading the equation as x^2 - 6x +9 - We've flipped the values of a and b in the equation a^2 - 2ab + b^2). Based on my approach |x - 3| would be the correct answer and on yours |3 - x| is the correct answer.

Does my above para make any sense? If it does, and if the question has both of these options, which would be the correct answer? If it doesn't please help me understand my mistake.
_________________

Cheers!!

JA If you like my post, let me know. Give me a kudos!

I got the roots for the equation as (x-3)^2 (reading the equation as x^2 - 6x +9 - We've flipped the values of a and b in the equation a^2 - 2ab + b^2). Based on my approach |x - 3| would be the correct answer and on yours |3 - x| is the correct answer.

Does my above para make any sense? If it does, and if the question has both of these options, which would be the correct answer? If it doesn't please help me understand my mistake.

The point is that |x - 3| = |3 - x|. Both indicate the distance between x and 3.
_________________

Which of the following always equals \(\sqrt{9 + x^2 - 6x}\) ?

A. \(x - 3\) B. \(3 + x\) C. \(|3 - x|\) D. \(|3 + x|\) E. \(3 - x\)

\(\sqrt{9 + x^2 - 6x} = \sqrt{(3 - x)^2} = |3 - x|\) (by the definition of a square root).

Answer: C

i answered choice a : x-3

i took the the expression as x^2-6x+9 ---> (x-3)^2. no my understanding is that on GMAT when some expression is under root sign, only positive root is considered...? is it correct. if it is so, the we need not take mod value...? pls help

Which of the following always equals \(\sqrt{9 + x^2 - 6x}\) ?

A. \(x - 3\) B. \(3 + x\) C. \(|3 - x|\) D. \(|3 + x|\) E. \(3 - x\)

\(\sqrt{9 + x^2 - 6x} = \sqrt{(3 - x)^2} = |3 - x|\) (by the definition of a square root).

Answer: C

i answered choice a : x-3

i took the the expression as x^2-6x+9 ---> (x-3)^2. no my understanding is that on GMAT when some expression is under root sign, only positive root is considered...? is it correct. if it is so, the we need not take mod value...? pls help

Exactly because the square root function cannot give negative result the answer is |3-x|, an absolute value of 3-x, which also cannot be negative.

MUST KNOW: \(\sqrt{x^2}=|x|\):

The point here is that since square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\).

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: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

I'd advice to go through basics and only after to practice questions:

After solving this question I got the result as |X-3| and I answered it as X-3, And got this wrong. Thanks for giving the link which clears the basic of Absolute value including the result:

|A-B| = |B-A|

Which is clearly applicable in this question provided options.

Which of the following always equals \(\sqrt{9 + x^2 - 6x}\) ?

A. \(x - 3\) B. \(3 + x\) C. \(|3 - x|\) D. \(|3 + x|\) E. \(3 - x\)

\(\sqrt{9 + x^2 - 6x} = \sqrt{(3 - x)^2} = |3 - x|\) (by the definition of a square root).

Answer: C

i answered choice a : x-3

i took the the expression as x^2-6x+9 ---> (x-3)^2. no my understanding is that on GMAT when some expression is under root sign, only positive root is considered...? is it correct. if it is so, the we need not take mod value...? pls help

Exactly because the square root function cannot give negative result the answer is |3-x|, an absolute value of 3-x, which also cannot be negative.

MUST KNOW: \(\sqrt{x^2}=|x|\):

The point here is that since square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\).

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: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

I'd advice to go through basics and only after to practice questions:

Hello Bunuel, Vyshak I thought the equation was as follows: \(\sqrt{x^2}=x\), if \(x>{0}\); \(\sqrt{x^2}=-x\), if \(x<={0}\). Am i wrong? Please correct me if so. Thanks

Hello Bunuel, Vyshak I thought the equation was as follows: \(\sqrt{x^2}=x\), if \(x>{0}\); \(\sqrt{x^2}=-x\), if \(x<={0}\). Am i wrong? Please correct me if so. Thanks

You can put = sign in any of the two because \(\sqrt{0^2}=0=-0\).
_________________

Hello Bunuel, Vyshak I thought the equation was as follows: \(\sqrt{x^2}=x\), if \(x>{0}\); \(\sqrt{x^2}=-x\), if \(x<={0}\). Am i wrong? Please correct me if so. Thanks

You can put = sign in any of the two because \(\sqrt{0^2}=0=-0\).