Question

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

Answer 1

Let $f\left(x\right)=a{x}^2+bx+6$
As $a{x}^2+bx+6=0$ does not have distinct real roots, so
$D\le 0$
Putting $x=0$
$f\left(0\right)=6>0$
So,
$f\left(x\right)\ge 0,\forall x\in R$

Now,
$f\left(3\right)\ge 09a+3b+6\ge 0\Rightarrow 3a+b\ge -2$

Therefore, the least value of $3a+b$ is $-2.$
Hence, $|k|=2$