Exponents/Roots: Given that x is an integer greater than 1,

[II]

Given that x is an integer greater than 1, how can you determine whether the following expression CAN be an integer ?

X^1/4 + x^1/2

whats the best way to answer this ?

Thanks.
II

[parore26]

Determine that x has a 4th root. if x = 16, the answer of that expression = 2 + 4 = 6.

If x was 4, then the value of the expression wouldn't be an integer. Hope this helps.

[II]

Thanks Parore ... thats the way I did it ... was wondering it there was another way.

[netigen]

The expression evaluates to

(1+x^2) / x^4

This expression can never be an interger when x!=1

[Stuart@KaplanGMAT]

The expression evaluates to

(1+x^2) / x^4

This expression can never be an interger when x!=1

How did you arrive at that simplification? I ask because it's definitely not the same as the original expression.

The original expression (X^1/4 + x^1/2) will be an integer whenever x is a perfect quardic (i.e. has a 4th root that's an integer, for example, 1^4, 2^4, 3^4, 4^4, ...).

[netigen]

You are correct, I misread the original expression.

The expression evaluates to

(1+x^2) / x^4

This expression can never be an interger when x!=1

How did you arrive at that simplification? I ask because it's definitely not the same as the original expression.

The original expression (X^1/4 + x^1/2) will be an integer whenever x is a perfect quardic (i.e. has a 4th root that's an integer, for example, 1^4, 2^4, 3^4, 4^4, ...).