cschmidlapp Wrote:If x is positive, is x>3?
1) (x-1)^2 > 4
2) (x-2)^2 > 9
I got this question right but I wanted to confirm/check my reasoning. For statement one I took the square root of each side and got (x-1)>absolute value of 2. Adding one to both + and negative values of 2 leads to x>3 and x>-1. Because in question statement they tell you x is positive, and because x>3 is the most restrictive in this case, x has to be greater than 3 and cannot be 1 or 2. I used the same logic for statement 2 to get x>5 and x>-1, and answer is D, both statements are sufficient.
Well... not totally.
First, you're writing the absolute values on the wrong side. This may just be because you're using words rather than math symbols, but the correct versions are |x - 1| > 2 and |x - 2| > 3, respectively.
Not x - 1 > |2| and x - 2 > |3| (which is what your words literally say -- don't know whether you meant that or not.) If you had these, then the absolute-value symbols don't even do anything, because |2| and |3| are just 2 and 3.
--
More importantly, you've got the signs switched, and you've got "and" where you should have "or".
The solution to |x - 1| > 2 is
x - 1 > 2
OR x - 1
< -2
--> x > 3
OR x
< -1
Likewise, the solution to |x - 2| > 3 is x > 5
OR x
< -1.