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Is sqrt(7x) an integer?

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by ank_1 » Mon Feb 13, 2012 9:42 pm
If square root of 28x is an integer, is square root of 7x an integer?

My Reasoning: 1> i would write 28x as 2^2*7x. Therefore sqrt(28x) to be integer x has to be 7*n*n. When i substitute value of x in equation sqrt(7x), i will get an integer.

But the answer on one of the test says it cannot be determined by this statement.
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by pemdas » Mon Feb 13, 2012 9:58 pm
ank_1 wrote:If square root of 28x is an integer, is square root of 7x an integer?

My Reasoning: 1> i would write 28x as 2^2*7x. Therefore sqrt(28x) to be integer x has to be 7*n*n. When i substitute value of x in equation sqrt(7x), i will get an integer.

But the answer on one of the test says it cannot be determined by this statement.
Sqrt(28x)=integer. Is Sqrt(7x)=integer?
Sqrt(28x)=Sqrt(2*2*7*x)=2*Sqrt(7x)

Thus, Sqrt(28x) must be an integer, only if Sqrt(7x) is an integer. Answer is Yes.
Last edited by pemdas on Mon Feb 13, 2012 11:00 pm, edited 1 time in total.
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by Anurag@Gurome » Mon Feb 13, 2012 10:01 pm
ank_1 wrote:If square root of 28x is an integer, is square root of 7x an integer?

My Reasoning: 1> i would write 28x as 2^2*7x. Therefore sqrt(28x) to be integer x has to be 7*n*n. When i substitute value of x in equation sqrt(7x), i will get an integer.

But the answer on one of the test says it cannot be determined by this statement.
If √(28x) = integer, then is √(7x) an integer?
√(28x) = 2√(7x)
So, for 2√(7x) to be an integer, √(7x) should also be an integer. So, this is sufficient and the answer to the question is "yes".
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by ank_1 » Mon Feb 13, 2012 10:20 pm
@PEMDAS - sqrt(28x) cannot be 2sqrt(14x).

I agree with anurag.This question was on Kaplan test and it says the statement is insufficient.
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by pemdas » Mon Feb 13, 2012 10:56 pm
omg, i didn't know which question you put in pm

again, that was a blunder, of course Sqrt(28x) doesn't fall into the written above. I will update it.thanks for correction.
ank_1 wrote:@PEMDAS - sqrt(28x) cannot be 2sqrt(14x).

I agree with anurag.This question was on Kaplan test and it says the statement is insufficient.
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by bubbliiiiiiii » Thu Feb 16, 2012 12:09 am
Anurag@Gurome wrote:
ank_1 wrote:If square root of 28x is an integer, is square root of 7x an integer?

My Reasoning: 1> i would write 28x as 2^2*7x. Therefore sqrt(28x) to be integer x has to be 7*n*n. When i substitute value of x in equation sqrt(7x), i will get an integer.

But the answer on one of the test says it cannot be determined by this statement.
If √(28x) = integer, then is √(7x) an integer?
√(28x) = 2√(7x)
So, for 2√(7x) to be an integer, √(7x) should also be an integer. So, this is sufficient and the answer to the question is "yes".
Hi Sir,

As per you post, I understand that since 2sqrt(7x) is an integer, sqrt(7x) has to be an integer.

How about sqrt(7x) having some non-integer value, for instance 3.5, then 2sqrt(7x) results to 7 which is an integer.

Please share your thoughts and help me understand what am I missing in evaluating the problem?
Regards,

Pranay
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