neiln413b wrote:

The integers x and y are positive. Is the \(\sqrt{x^2 + 7xy}\) an integer?

(1) y = 5x

(2) y - x = 4

Square root of x^2 + 7xy will be an integer if x^2 + 7xy is a perfect square.

(1) Substituting y=5x, we get

x^2 + 7xy = x^2 + 7*x*5x = x^2 + 35x^2 = 36x^2.

Now since x is a positive integer, x^2 is a perfect square, and 36 is already a perfect square. So 36x^2 will also be a perfect square. Thus its square root will be an integer. Sufficient.

(2) Substituting y = 4+x we get

x^2 + 7xy = x^2 + 7x(4+x) = x^2 + 28x + 7x^2 = 8x^2 + 28x = 4x(2x+7).

Now say x=1, then 4x(2x+7) = 4*1(2+7) = 4*9 which is a perfect square.

But if say x=2, then 4x(2x+7) = 4*2(4+7) = 8*11 which is not a perfect square.

Thus this statement is Not Sufficient.

Hence

A answer