Question

The value of p for which the quadratic equation is x(x-4) + p = 0 has real roots, is :

Answer 1

x(x – 4) + p = 0

⇒ x² – 4x + p = 0

Comparing this with standard form of quadratic equation ax² + bx + c = 0,

We get a = 1, b = – 4 and c = p

Discriminant (D) = b² – 4ac = (– 4)² – 4(1) (p) = 16 – 4p

For real roots, D ≥ 0

⇒ 16 – 4p ≥ 0

⇒ 16 ≥ 4p

⇒ 4 ≥ p i.e., p ≤ 4

So, in order to have real roots, p ≥ 4.