If we remove the exponents from the left, then we need can also show the radical of 4 and 9 (right side) therefore we get

a) x-1 > 2 therefore x > 2+1 or x>3 - sufficient
b) x-2 > 3 therefore x > 3+2 or x>5 - sufficient

Answer is D

I am not sure if my method to solve is correct, because inequalities are not my strongest point. So would like to understand from other members if this is correct. I would have done this on the test as well.

Thanks

1. (x-1)^2 > 4
If we taking sq root on both the side then there are two possible equations:
(x-1) > 2 --- (a)
-(x-1) > 2 --- (b)

Take (a)
X > 3 --- (c)

Take (b)
-(x-1) > 2
(X-1) < - 2
X< -1 – (d)
But, x is positive. Hence, only (c) is possible.
Hence, 1 is sufficient.

2. Similarly, second input will give us two domains
x>5 or x< -1. But, as x is positive, only x>5 is possible. Hence, 2 is also sufficient.

IMO D.

Bhattu wrote:If we remove the exponents from the left, then we need can also show the radical of 4 and 9 (right side) therefore we get

a) x-1 > 2 therefore x > 2+1 or x>3 - sufficient
b) x-2 > 3 therefore x > 3+2 or x>5 - sufficient

Answer is D

I am not sure if my method to solve is correct, because inequalities are not my strongest point. So would like to understand from other members if this is correct. I would have done this on the test as well.

Thanks

Each equation can be written as
lx-1l>2
x-1>2 or -(x-1)>2
x>3 or x-1<-2
x>3 or x<-1

since it is given that x is positive so x<-1 is NA-- suff


Similarily for B

lx-2l>3
x-2>3 or -(x-2)>3
x>5 or x-2<-3
x>5 or x<-1
since it is given that x is positive so x<-1 is NA-- suff

Easy one it has to be D