Radical DS and response from veritas prep professor

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I was hoping you could explain this problem for me? I got it from the beatthegmat.com practice bank. I'm confused, especially because I don't see where they get that 0 (1. Premise second slide) is positive integer x. I know that 0 is even but I thought it was not positive nor negative. In the third slide they change x to 4. Can you help on this problem?

You are right that 0 is even but that it is neither positive nor negative, and you're right that they shouldn't be using x=0 as a possible case because it violates the condition they've specified. Typo on BTG! But Statement 1 is still insufficient because y could be 0 (since we're not told that y is positive), so if you let x be any perfect square and let y be 0, you will satisfy Statement 1 and get a YES answer to the question; if, on the other hand, you let x and y be different things (like 1 and 3, for example, or 2 and 7), then you'll satisfy Statement 1 and get a NO answer. So Statement 1 gives you a maybe.

Statement 2 gives you a YES, DEFINITELY, because we can transform Statement 2 into 8x^2 = y^2, and then when we substitute 8x^2 in for y^2 in the original question, the question winds up saying "Is root(x^2 + 8x^2) an integer?"... in other words, is root(9x^2) an integer?...in other words, is 3x an integer? And yes, it is, since x itself is an integer.




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by amit2k9 » Tue Jul 05, 2011 9:57 pm
a x=1. (1+3)^1/2 = 2 but (1+9)^1/2 = 10^1/2 = not integer (3.23 approx)
and for x=1 y=0

(1+0)^1/2 = integer and (1^2 + 0^2)1/2 = integer. not sufficient.


b 8x^2 = y^2 means (9x^2)1/2 = 3(x^2)^1/2 will be an integer.

sufficient.
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