If p ≠ 2 in (p − 2)x² − 2(p + 2)x + (p + 6) = 0, what is the condition on p for real roots?

Correct answer: Every p ≠ 2

The governing concept is that a quadratic has real roots when its discriminant is non-negative. Here a = p − 2, b = −2(p + 2), and c = p + 6. Compute D = [−2(p + 2)]² − 4(p − 2)(p + 6) = 4(p + 2)² − 4(p² + 4p − 12) = 4(p² + 4p + 4 − p² − 4p + 12) = 64. This calculation shows D is always positive, not 40 − 8p. Therefore, for every p ≠ 2, the equation is genuinely quadratic and has two real distinct roots. None of the supplied options states this correctly: A unnecessarily restricts p, B is opposite, C violates p ≠ 2, and D is the only option that says every p ≠ 2. Hence option D, not the supplied key A, is correct.