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In the following determine the set of value of P for which the given quadratic equation has real roots. Px2 + 4x + 1 = 0
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- P ≠ 4
- P > 4
- P ≤ 4
- P ≥ 4
Correct Option: C
Given :- Px2 + 4x + 1 = 0
Compare with Ax 2 + Bx + C = 0, we get
A = P, B = 4, C = 1
For real roots,
B2 - 4AC ≥ 0
⇒ 16 - 4P ≥ 0
⇒ 16 ≥ 4P
⇒ P ≤ 16/4 ⇒ P ≤ 4.
Hence , required answer is option C .