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or Sqrt(4X)=2Sqrt(x), if this is an integer then Sqrt(x) is either an integer or is equal to 1/2 or Y/2....in that case its not an integer... again I am assuming irrational numbers can be viewed as fractions....there is no precise defintion that they are not...

or Sqrt(4X)=2Sqrt(x), if this is an integer then Sqrt(x) is either an integer or is equal to 1/2 or Y/2....in that case its not an integer... again I am assuming irrational numbers can be viewed as fractions....there is no precise defintion that they are not...

I would go with E on this...

Shud be "A"...X is +ve integer can't be 1/4.

Saurya, u r correct in ur approach except that ur variable K has to be an even integer as the LHS has a factor of 2.

or Sqrt(4X)=2Sqrt(x), if this is an integer then Sqrt(x) is either an integer or is equal to 1/2 or Y/2....in that case its not an integer... again I am assuming irrational numbers can be viewed as fractions....there is no precise defintion that they are not...

I would go with E on this...

Shud be "A"...X is +ve integer can't be 1/4.

Saurya, u r correct in ur approach except that ur variable K has to be an even integer as the LHS has a factor of 2.

So baner, what is wrong with my approach. So, answer should be E with my way as we can't say if k is odd or even?
Can you plz explain this.
S _________________

Himalaya, can you know what is conceptual error in my method? Thanks mate

saurya_s wrote:

If x is a positive integr is x^(1/2) (Sq root of x) is an integer? 1. (4x)^(1/2) is an intger 2. (3x)^(1/2) is not integer. from 1, let (4x)^(1/2) =k where k is an integer. x^(1/2)=k/2. so does value of root x depends on odd/even. What wrong am I doing. What is the way to approach this sort of problem.

surya, it doesnot matter. if (4x)^(1/2) is an intger, x^(1/2) must be an intger because sqrt(4) is an integer (+or-2).

you did x^(1/2)=k/2, which still must give you an integer. if you are supposing x as fraction (e.g1/4 or 0.25), which is not possible in this case because x is a positive integer as per the statement. so sqrt(x) must have an integer.