Hi Bunuel,

In ques - Link: https://gmatclub.com/forum/d01-183474.html#p1571995

You have made a point that:
"When the GMAT provides the square root sign for an even root, such as a square root, fourth root, erc. then the only accepted answer is the positive root.

That is:
√9=3, NOT +3 or -3;

Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and -3."


Why same is not valid here?
Why are we considering the value of sqrt as positive and negative?

Where do you see a contradiction? When we have that \(\sqrt{x^2}=|x|\), still the square root is giving a non-negative result because the absolute value of a number is always non-negative. Please read the previous pages, I think several posts there explain this clearly.
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Fab Question!

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation.

Good question.

Loved the question. It really put my mind to work

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and the explanation isn't clear enough, please elaborate.

I had trouble understanding the solution at first but then by looking at different cases I think I understood, please let me know if below is right:
we have the simplified equation as |x−3|+√(2−x)+x−3
as "each expression under the square root is greater than or equal to 0" then (2−x) >= 0 which means x=<2
then looking at x−3 this means it will alway be a negative whereas |x−3| will be the same value but positive, therefore they will cancel each other and the equation will be equal to √(2−x)

Correct. One clarification: \(x \le 2\) means x-3 is negative, so \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).
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