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Sec 3, #25
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cindyqtran
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mschling52
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I agree with B for this one.
Statement (1) gives us that a^2+b^2 is an integer, but that may or may not mean that the sqrt is an integer.
However, Statement (2) tells us that a^2 = 3(b^2). Then we can substitute this into the original equation
sqrt(a^2+b^2) = sqrt(3(b^2)+(b^2)) = sqrt(4(b^2)) = 2b
Since we are given that b is an integer, 2b will also be an integer. Hence, sqrt(a^2+b^2) = 2b is an integer.
Statement (1) gives us that a^2+b^2 is an integer, but that may or may not mean that the sqrt is an integer.
However, Statement (2) tells us that a^2 = 3(b^2). Then we can substitute this into the original equation
sqrt(a^2+b^2) = sqrt(3(b^2)+(b^2)) = sqrt(4(b^2)) = 2b
Since we are given that b is an integer, 2b will also be an integer. Hence, sqrt(a^2+b^2) = 2b is an integer.
