Multiple choice

The values of P for which the given quadratic equation has real roots, $P{x}^{2}+4x+1=0$ is

  1. $P<4$
  2. $P\leq 4$
  3. $P>4$
  4. $P\geq 4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For real roots, discriminant D >= 0. D = b^2 - 4ac = 4^2 - 4(P)(1) = 16 - 4P. 16 - 4P >= 0 implies 16 >= 4P, so P <= 4.

AI explanation

For the quadratic equation Px^2 + 4x + 1 = 0 to have real roots, the discriminant must be greater than or equal to zero, so D = 4^2 - 4(P)(1) >= 0. Simplifying 16 - 4P >= 0 gives 16 >= 4P. Dividing by 4 yields P <= 4.

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