Question

If the equation ax² + bx + 6 = 0 does not have two distinct real roots, then the least value of 3a + b is

(A)  3 

(B)  −3 

(C)  2 

(D)  −2

Answer 1

Correct option  (D) −2

f(x) = ax² + bx + 6 = 0

f(0) = 6 is positive. Therefore,

f(x) = ax² + bx + 6 ≥ 0   ∀ ∈x R

⇒ f(3) = 9a + 3b + 6 ≥ 0

⇒ 3(3a + b) ≥ −6

⇒ 3a + b ≥ −2