Question

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

• A. $1$
• B. $-1$
• C. $-2$
• D. $2$

Answer 1

The correct option is C $-2$

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

Hence, the least value of $3a+b$ is $-2.$