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.
Aptitude
Quadratic Equation
Difficulty: Easy
Choose an option
-
AP ≠ 4
-
BP > 4
-
CP ≤ 4
-
DP ≥ 4
Answer
Correct Answer: P ≤ 4
Explanation
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:
Δ = 16 − 4P.Require Δ ≥ 0 ⇒ 16 − 4P ≥ 0 ⇒ −4P ≥ −16.Multiply by −1 and reverse inequality: 4P ≤ 16 ⇒ P ≤ 4.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:
P ≤ 4