If $ax^2+bx+6=0$ doesn't have $2$ distinct real roots, then find the least value of $ (3a+b)$

Question

If $ax^2+bx+6=0$ doesn't have $2$ distinct real roots, then find the least value of $(3a+b)$ $a,b\in \mathbb R$

Any hint for this question?

Answer 1

Picking up from $b^2 \leq 24a$, we have $b^2 + 8b \leq 8(3a + b)$ or $$\frac{1}{8}\left( b^2 + 8b \right) \leq 3a + b$$

Now what is the minimum value of the LHS?

Answer 2

HINT: What happens when $D=b^2-4ac>0$ and what happens when $D=0$ and $D<0$