Question

If ax² + bx + 8 = 0 does not have 2 distinct real roots, then find the minimum value of 2a + b.

Answer 1

Let f(x) = ax² + bx + 8.

Since f(x) = 0 does not have 2 distinct real roots, hence we have either f(x) ≥ 0 ∀ x ∈ R or f (x) ≤ 0 ∀ x ∈ R.

Since f(0) = 8, therefore

⇒ f(x) ≥ 0 ∀ x ∈ R

In particular, f(2) ≥ 0

⇒ 4a + 2b + 8 ≥ 0

⇒ 2a + b ≥ − 4

Hence, the minimum value is −4.