$ax^2 + bx +6$ does not have two distinct real roots , then what will be the least value of $3a + b$?
I know that $D$ will be less than or equal to. But least value of $3a + b$ can not be deduced from that.
Can anyone please help me?
[Math] $ ax^2 + bx +6 $ does not have two distinct real roots , then what will be the least value of $ 3a + b $
quadratics
Best Answer
Guide:
We want to minimize $3a+b$ subject to $b^2-24a \le 0$.
When the minimal is attained, $b^2=24a$ is satisfied. That is $a=\frac{b^2}{24}$.
Hence, we want to minimize $\frac{3b^2}{24}+b=\frac{b^2}{8}+b$, can you take it on from here? Say, by using calculus or completing the square.
Edit:
Relevant Desmos link. As we change the value of $k$, notice that the optimal value must touches the boundary.