This solution is more on the logical side than algebric. I am sure other s may have an algebric solution

x is positive x can be a positive fraction or a postive integer

Stmt I

(x-1) ^2>4 only possible if x>3 (cant be a positive fraction less than 3 or a pos intger less than 3)

SUFF

Stmt II
(x-2 ^2>9 only possible if x>3 (cant be a positive fraction less than 3 or a pos intger less than 3)

suff


D)

If X is positive, is X > 3.

Statement 1: (X-1)^2 > 4
Statement 2: (X-2)^2 > 9

The given information that X is positive is necessary to solve this problem correctly.

Statement 1:
(X-1)^2 > 4 = X-1 > 2 or X - 1 < -2, so x > 3 or X < -1. However we know that X has to be positive, so ignore the possibility that X < -1. So we know that X > 3. (Sufficient)

Statement 2:
(X-2)^2 > 9 = X-2 > 3 or X - 2 < -3, so X > 5 or X < -1. However, once again, we know that X has to be positive, so ignore the possibility that X < -1. So we know that X > 5 (and therefore X > 3). (Sufficient).

Both statements are sufficient, so the answer is D.

(X-1)^2 > 4 = X-1 > +/- 2,

RipRop,
I may be missing something here. Should this not be

-(x-1) >2

Then u get x<-1

In your case if x-1>-2
x>-1

Did I miss something here?

Thanks, you're right. I was lazy with that part. The solutions were right but the way I displayed the formula was incocrect. I fixed it above for the OP.

Thanks RipRop!

This is my way for getting to the 2 solutions (explaining it just for FYI purposes)

(x-1) ^ 2 > 4
can be
(x-1) ^ 2 > 2 ^ 2 (case 1)
or (x-1) ^ 2 > (-2) ^ 2 case 2


Case 1:

(x-1) ^ 2 > 2 ^ 2
sqrt((x-1)^2) > sqrt (2^2)

|x-1| > |2| (definition of sqrt(x^2) = |x| |2| is 2)

x-1>2
x>3

-(x-1)>2
x<-1

case 2:

(x-1) ^ 2 > (-2) ^ 2
sqrt((x-1)^2) > sqrt ((-2)^2)

|x-1| > |-2| (definition of sqrt(x^2) = |x| |-2| is 2)

x-1>2

x>3

-(x-1)>2
x<-1

Comments/Critiques/Suggestions welcome!

Thanks Cramya for the clear and wonderful explanation !!!

cramya wrote:Thanks RipRop!

This is my way for getting to the 2 solutions (explaining it just for FYI purposes)

(x-1) ^ 2 > 4
can be
(x-1) ^ 2 > 2 ^ 2 (case 1)
or (x-1) ^ 2 > (-2) ^ 2 case 2


Case 1:

(x-1) ^ 2 > 2 ^ 2
sqrt((x-1)^2) > sqrt (2^2)

|x-1| > |2| (definition of sqrt(x^2) = |x| |2| is 2)

x-1>2
x>3

-(x-1)>2
x<-1

case 2:

(x-1) ^ 2 > (-2) ^ 2
sqrt((x-1)^2) > sqrt ((-2)^2)

|x-1| > |-2| (definition of sqrt(x^2) = |x| |-2| is 2)

x-1>2

x>3

-(x-1)>2
x<-1

Comments/Critiques/Suggestions welcome!

Cramya,
would like to suggest a similar yet simple approach. May be you already know this.

given from 1
(x-1)^2>4
or |x-1| > 2
1 is the critical point whr x-1=0,
the distance 2 units greater from 1 on both sides.
it gives x<-1 or x>3 but we have x>0 so, x<-1 is ignored.
similarly
(x-2)^2 > 9
|x-2| > 3
2 is the critical point.
the distance is 3 units greater from 2 on both sides.
it gives x<-1 or x>5
again we can ignore x<-1 as x is positive.

Hence ans is D.
Suggestions welcome.