Kshah001 wrote:
Experts Please suggest if this method is correct. I solved it as below. Request Bunuel, Veritaskarishma to suggest.
Also I am confused - why am i getting different range for x. Please help
St1 (x-1)^2>4
x^2-2x+1>4
x^2-2x>3
x(x-2)>3
So x>3 or x-2>3
X>3 or x>5 Hence St 1 sufficient
St 2 (x-2)^2>9
x^2-4x+4>9
x^2-4x>5
x(x-4)>5
x>5 or x-4>5
x>5 or x>9
st 2 sufficient.
Ans Choice D
Your mistake is in this step:
"x(x-2)>3
So x>3 or x-2>3"
You are most likely confusing this with when it is equal to zero -- for example:
x(x-2) = 0 ---> in this case, x = 0 OR x-2 = 0. (One of the two things we are multiplying on the left must be zero, to get zero on the right)
Important ---> This does not apply if it's not equal to zero.
Instead, we want to take the square root of both sides:
(1) \((x - 1)^2\) > 4
|x-1| > 2 (Important: when we take the square root of a square, we get the absolute value. It's a common mistake to forget this. For this problem, it doesn't affect the answer, because the question says "if x is positive", and the negative values don't apply. Another common mistake is to miss the "If x is positive" -- read carefully!)
Case A: If x-1 is positive, then x-1 > 2 --->
x > 3Case B: If x-1 is negative, then the absolute value flips the signs on the left: -x+1 > 2 ---> x < -1 ---> this doesn't apply, because the question says "If x is positive".
Therefore, our answer to the question, "is x>3?", is YES, and
(1) is sufficient.We do the same process for statement 2:
(2) \((x - 2)^2\) > 9
|x-2| > 3
x > 5 Again, the negative case doesn't apply here. (If x-2 is negative, then the absolute value flips the signs on the left: -x+2 > 3 ---> x < -1 ---> this doesn't apply, because the question says "If x is positive")
Therefore, our answer to the question, "is x>3?", is YES, and
(2) is sufficient.Answer is
D _________________
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