Real-root condition by discriminant: For Px^2 + 4x + 1 = 0, determine the set of all real values of P for which the quadratic has real roots.

Introduction / Context:
A quadratic az^2 + bz + c has real roots if and only if its discriminant Δ = b^2 − 4ac is nonnegative. We apply this directly to Px^2 + 4x + 1 = 0, treating P as a parameter.

Given Data / Assumptions:

• a = P, b = 4, c = 1.
• P is real (and can be zero; then the equation is linear).

Concept / Approach:
Impose Δ ≥ 0 ⇒ 4^2 − 4*P*1 ≥ 0 ⇒ 16 − 4P ≥ 0. Solve the inequality for P.

Step-by-Step Solution:

Verification / Alternative check:
At P = 4, Δ = 0, giving a repeated real root. For any P < 4, Δ > 0 gives two distinct real roots. For P > 4, Δ < 0 yields complex roots.

Why Other Options Are Wrong:

• P ≠ 4 or P ≥ 4 or P > 4: These contradict Δ ≥ 0.

Common Pitfalls:
Forgetting to reverse the inequality after multiplying by −1, or thinking that a = 0 forbids roots (it simply becomes linear with one real root).

Final Answer: