jabhatta2
AndrewN
banksy
If b is an integer, is \(\sqrt {a^2+b^2}\) an integer?
(1) a^2 + b^2 is an integer.
(2) a^2 – 3b^2 = 0
DS46402.01
NUMBER PROPERTY CONCEPT: This may be obvious to most people on this forum, but since no one has touched on it in the thread and the question is missed by about a quarter of the people who attempt it, I thought I would draw attention to the fact that
\(\sqrt{0}=0\)That is,
even if a and b are 0 in the above expression, √0 is still 0, an integer, and is defined, so the answer cannot be (E).
- Andrew
Hi
AndrewN - could you please elaborate a bit more about why you say the answer CANNOT be (E), based on the pink ?
I agree with the concept mentioned in the pink above but if a^2+b^2 was say 5...then √5 WOULD NOT GIVEN YOU an integer.
So you could get a Yes/ No, making E a contender
Hello,
jabhatta2. Could this be a case in which you may have overlooked some information in the problem, namely that
b is an integer? I agree with you that, based on Statement (1) alone, the number under the radical could be 5, or it could be some perfect square. However, as others have worked out above, when you consider Statement (2) alone, you can see that
a can be substituted out for its equivalent expression in
b:
\(a^2 - 3b^2 = 0\)
\(a^2 = 3b^2\)
\(\sqrt {(3b^2)+b^2}\)
\(\sqrt {4b^2}\)
The 4 can be factored into two 2's and placed on the outside of the radical, leaving
\(2\sqrt {b^2}\)
Since we know that, again,
b is an integer, the value under the radical CANNOT be 5. I was pointing out that the smallest value it can be is 0, and that √0 still leads to an integer value as an answer. Thus, (B) must be the answer (and (E), by extension, cannot).
- Andrew