Assume y = sqrtx

Then you can rewrite the equations to:

1) y^2-2*sqrt3*y = 3
y^2-2*sqrt3*y-3 = 0
(y-sqrt3)(y-sqrt3) = 0
y = sqrt3
x = sqrt3^2 = 3

You can back solve to the original here to check:
3 - 2(sqrt(3*3) = -3
3 - 2(sqrt9) = -3
3 - 2(3) = -3
3 - 6 = -3

1 is sufficient

2) y^2 - sqrt3*y - sqrt5*y - sqrt15 = 0
(y-sqrt3)(y-sqrt5) = 0
y=sqrt3 OR y = sqrt5
X = 3 or x = 5

Back solve just to prove it
3 - sqrt(3*3) - sqrt (3*5) + sqrt15 = 0
3 - 3 - sqrt15 + sqrt15 = 0
0 =0
AND
5 - sqrt(3*5) - sqrt(5*5) + sqrt15 = 0
5 - sqrt15 - 5 + sqrt15 = 0
0=0

2 insufficient.

Answer = A

-Carrie

That was evil and took very very long to solve. But thanks as it gives me one more way to look at square roots that might be very helpful on the test.