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

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by II » Fri Feb 22, 2008 1:00 pm
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.
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Last edited by II on Mon May 05, 2008 2:28 am, edited 1 time in total.
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by parore26 » Fri Feb 22, 2008 2:09 pm
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.
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by II » Fri Feb 22, 2008 3:02 pm
Thanks Parore ... thats the way I did it ... was wondering it there was another way.
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by netigen » Fri Feb 22, 2008 3:40 pm
The expression evaluates to

(1+x^2) / x^4

This expression can never be an interger when x!=1
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by Stuart@KaplanGMAT » Fri Feb 22, 2008 4:22 pm
netigen wrote: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, ...).
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by netigen » Fri Feb 22, 2008 4:47 pm
You are correct, I misread the original expression.
Stuart Kovinsky wrote:
netigen wrote: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, ...).
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