If x > 1, is [m][square_root]{3x}[/square_root][/m] an integer? (a) s

If x > 1, is \(\sqrt{{3x}}\) an integer?


(a) \(\sqrt{\frac{x}{3}}\)is not an integer
If \(x=\frac{4}{3}\), then yes.
If \(x=4\), then no.
Insufficient

(b) \( x ÷\frac{1}{\sqrt{3}}\) is an integer

\( x ÷\frac{1}{\sqrt{3}}=x\sqrt{3}\)
So x has to be some perfect square *\(\sqrt{3}\)
Let \(x=a^2\sqrt{3}\), where a is an integer.
\(\sqrt{3x}=\sqrt{3*a^2*\sqrt{3}}=a\sqrt{\sqrt{3}}\)
Answer is always no.
Sufficient

B
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We may also ask "Is 3x a square?"

In order to have \(\sqrt{3x} = i\), we need x in the form of \(\frac{i^2}{3}\), where \(i\) is a positive integer.

Statement 1:

We may let \(x = \frac{2^2}{3}\). Then \(\sqrt{3x}\) would be an integer. Insufficient.

Statement 2:

\(\sqrt{3}x = j\), so we have \(x = \frac{\sqrt{3}j}{3}\). Compare \(\sqrt{3}j\) and \(i^2\), the former cannot be any integer squared so there is no case where \(\sqrt{3}j = i^2\). Then x cannot fit the criteria in any case, sufficient.

Ans: B
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How can x = 4/3 in the 1st statement? because if x=4/3, then sq. root of (x/3) will not be an integer... which violates statement 1

But that is exactly what the statement I tells.
Square root of x/3 is not an integer.
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