Minimum value of $4a+b$ Let $ax^2+bx+8=0$ be an equation which has no distinct real roots then what is the least value of $4a+b$ where $a,b\in \Bbb R$. 
My Try:
I differentiated the given function to get $f'(x)=2ax+b$ now if we put $x=2$ in this we get the required value . Now it's given that equation has no distinct roots so it can be a perfect square thus $b^2\leq 32a$. But now I don't know how to continue then. Also note that it's given that a,bare real so we cannot apply AM-GM inequality as they maybe negative. Thanks.
 A: This is the solution of DeepSea, just a bit more detailed. Since the constraint given in the question has to do with the roots of the polynomial, try solving it by completing the square (assuming $a > 0$)
$$
ax^2 + bx + 8 = 0 \iff x^2 + \frac ba x + \frac 8a = 0 \iff \left(x + \frac{b}{2a}\right)^2 + \frac 8a - \frac{b^2}{4a^2} = 0,
$$
and conclude that 
$$
x_{1,2} = -\frac{b}{2a} \pm \sqrt{\frac{b^2}{4a^2} - \frac 8a} = \frac{-b \pm \sqrt{b^2 - 32a}}{2a}.
$$
Note that if $b^2 - 32a = 0$ we get a double root, and if $b^2 - 32a < 0$ we get no real roots. Therefore for the polynomial to have no distinct real roots, we wish that $b^2 - 32a \leq 0$ or equivalently
$$
32a \geq b^2 \iff 4a \geq \frac{b^2}{8}.
$$
Adding $b$ to each side we get that
$$
4a + b \geq \frac{b^2}{8} + b = \frac{(b+4)^2 - 16}{8},
$$
where we once again completed the square to get the final expression. Realize that the least value of $4a + b$ is the minimal value of the right hand side, with $b \in \mathbb{R}$. Since $(b+4)^2 \geq 0$ for all values of $b$ conclude that the minimal value is $-16/8 = -2$.
A: We have: $\triangle \leq 0\implies b^2 - 32a \leq 0\implies 4a+b \geq \dfrac{b^2}{8} + b= \dfrac{b^2+8b}{8}= \dfrac{(b+4)^2-16}{8}\geq -2\implies \text{min} = -2$
