Since negative numbers do not have square roots,x²-4y² must be non-negative. Thus:

x²-4y²≥0

(x-2y)(x+2y)≥0

This can be true only if both (x-2y) and (x+2y) are non-negative or both are non-positive.

I. First we take the first case (both are non-negative).

Thus:

x-2y≥0 which means x≥2y

x+2y≥0 which means x≥-2y

Thus the solution set is x≥2y ∩ x≥-2y

II. Now we take the second case (both are non-positive)

Thus:

0≥x-2y which means 2y≥x

0≥x+2y which means -2y≥x

This can only be true when x=y=0, which is included in the solution set of case 1.

Thus the answer is x≥2y ∩ x≥-2y