Suppose 𝑌 is distributed uniformly in the open interval (1,6). The probability that the polynomial 3𝑥2 + 6𝑥𝑌 + 3𝑌 + 6 has only real roots is

Q. Suppose 𝑌 is distributed uniformly in the open interval (1,6). The probability that the polynomial 3𝑥² + 6𝑥𝑌 + 3𝑌 + 6 has only real roots is (rounded off to 1 decimal place)

Solution:

For a quadratic polynomial ax2 + bx + c = 0. There are three condition:

b2 – 4ac > 0 {real and distinct root, i.e., two real roots}

b2 – 4ac = 0 {real and equal roots, i.e., only one real root}

b2 – 4ac

Polynomial 3×2 + 6xY + 3Y + 6 has only real roots,

⇒ b2 – 4ax ≥ 0

⇒ (6Y)2 – 4(3) (3Y+ 6) ≥ 0

⇒ Y2 – Y + 2 ≥ 0

Y ∈ (–∞, – 1] ∩ [2, ∞)

⇒ Y ∈ [2, 6)

Since y is uniformly distributed in (1, 6).

Probability distributed function,

f(Y) = (1/5), 1 6

Hence