Which statement correctly solves (x^2\le 0) over real numbers?

Correct answer: (x=0)

The direct answer is A: x=0. For every real number x, its square satisfies x²≥0. Therefore x²≤0 can happen only when x² is exactly 0. A square is zero only when its base is zero, so x=0. A is correct. B, x<0, is wrong because a negative number has a positive square; for example, (−2)²=4, not a value at most zero. C, x>0, is wrong for the same reason: a positive number also has a positive square, for example 2²=4. D, all real x, is wrong because nonzero values produce positive squares; x=1 gives x²=1>0. The solution set contains exactly one point, zero. On a graph or number line, mark only 0 with a closed point because equality is included. Do not mistakenly think that “≤0” means all negative x; it describes the square’s value, not x itself. Memory cue: a real square never goes below zero, and it touches zero only at zero.