If x² − 2px + (p² − 64) = 0, what is the nature of the roots for any real value of p?

Correct answer: Two real, rational and distinct roots

The governing concept is the discriminant of a quadratic equation ax² + bx + c = 0. Its value is D = b² − 4ac. Here a = 1, b = −2p and c = p² − 64. Therefore, D = (−2p)² − 4(1)(p² − 64) = 4p² − 4p² + 256 = 256. This is positive for every real value of p, so the roots are real and distinct. Moreover, 256 = 16² is a perfect square, and the coefficients are real, so the roots are rational. In fact, the roots are p + 8 and p − 8. Thus option A is correct. Option B would require D = 0, option C would require D < 0, and option D is wrong because the discriminant is a perfect square.