This one might have been discussed before on this forum but I could not find it. Kindly explain.
OA is B
Gmat Prep - sqrt [(x-3)^2]
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- ssmiles08
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Can anyone explain this ques in detail???
https://www.beatthegmat.com/gmat-prep-ques-t37241.html
Similar problem should give you a better idea
- Domnu
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First of all,
( sqrt(x-3) )^2 = |x - 3|
Now, we need to look at |x - 3| = 3 - x. For the first part, x != 3 doesn't give us anything... if x = 2 they are equal, and for x = 10 they are not equal.
The second part is sufficient.. particularly, it says that x < 0. If this is the case, then |x - 3| = 3 - x, since the rhs is always going to be positive.
So, the answer is B.
( sqrt(x-3) )^2 = |x - 3|
Now, we need to look at |x - 3| = 3 - x. For the first part, x != 3 doesn't give us anything... if x = 2 they are equal, and for x = 10 they are not equal.
The second part is sufficient.. particularly, it says that x < 0. If this is the case, then |x - 3| = 3 - x, since the rhs is always going to be positive.
So, the answer is B.
Have you wondered how you could have found such a treasure? -T
"statement 2 tells you that x is negative..
If you take any negative value and substitute in the given equation you wil get LHS=RHS...
Thus B is sufficient"
Hey man, why is LHS always positive? A square root of a positive number can be a +ve or -ve isn't it?. I know that the square root of a negative number is complex.
If you take any negative value and substitute in the given equation you wil get LHS=RHS...
Thus B is sufficient"
Hey man, why is LHS always positive? A square root of a positive number can be a +ve or -ve isn't it?. I know that the square root of a negative number is complex.
200 or 800. It don't matter no more.