What is the correct statement about the real roots of x² + 2(2y−1)x + 4y² + 3 = 0?

Correct answer: It has real roots only when y ≤ −1/2

For x² + 2(2y−1)x + 4y² + 3 = 0, the coefficients are a = 1, b = 2(2y−1) and c = 4y² + 3. The discriminant is D = [2(2y−1)]² − 4(4y²+3) = 4(2y−1)² − 16y² − 12. Expanding gives 4(4y²−4y+1) − 16y² − 12 = −16y − 8 = −8(2y+1). Real roots require D ≥ 0, so −8(2y+1) ≥ 0, which gives y ≤ −1/2. Equality at y = −1/2 gives equal roots; smaller values give distinct real roots. Therefore option A is correct. The other statements incorrectly claim an unconditional result or restrict y to an incorrect single value.