BTGmoderatorDC wrote: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
OA B
Source: Official Guide
Let's take each statement one by one.
(1) a^2 + b^2 is an integer.
Case 1: Say a = √3 and b = 1, then \(\sqrt {a^2+b^2}= \sqrt {3+1}=2\), an integer. The answer is yes.
Case 2: Say a = 3 and b = 1, then \(\sqrt {a^2+b^2}= \sqrt {9+1}=√10\), not an integer. The answer is no.
No unique answer. Insufficient.
(2) a^2 - 3b^2 = 0
a^2 = 3b^2
Thus, \(\sqrt {a^2+b^2}= \sqrt {4b^2}=2b\), an integer. The answer is yes. Sufficient.
The correct answer:
B
Hope this helps!
-Jay
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