Question

If $a{x}^2+bx+6=0$ does not have distinct real roots, then the least value of $3a+b=$

Answer 1

Given equation $a{x}^2+bx+6=0$ does not have distinct real roots. Hence,
$\Rightarrow f\left(x\right)=a{x}^2+bx+6\le 0,\forall x\in R,\text{if}a<0$
or
$f\left(x\right)=a{x}^2+bx+6\ge 0,\forall x\in R,\text{if}a>0$

But $f\left(0\right)=6>0$
$\Rightarrow f\left(x\right)=a{x}^2+bx+6\ge 0,\forall x\in R$
$\Rightarrow f\left(3\right)=9a+3b+6\ge 0$
$\Rightarrow 3a+b\ge -2$
Therefore, the least value of $3a+b$ is $-2$.

Alternate Solution:
Given equation $a{x}^2+bx+6=0$ does not have distinct real roots.
Hence, $D\le 0$
$\Rightarrow b^2-24a\le 0$
$\Rightarrow 3a\ge \frac{b^2}{8}$

Now, $3a+b\ge \frac{b^2}{8}+b$
$\Rightarrow 3a+b\ge \frac{(b+4{)}^2-16}{8}$
$3a+b$ is minimum when $(b+4{)}^2$ is minimum.
Thererfore, minimum value of $3a+b=\frac{0-16}{8}=-2$