Question

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

• A. $4$
• B. $-1$
• C. $1$
• D. $-2$

Answer 1

Answer: (D)

$-2$

$a{x}^2+bx+6=0$
Substitute $b=k-3a$ to obtain $a{x}^2+(k-3a)x+6=0$
Since, the equation does not have real distinct roots.
Therefore, $D={(k-3a)}^2-24a\le 0$
$\Rightarrow 9a^2-6a(4+k)+k^2\le 0$
Above quadratic equation has real roots.
Therefore, $D=36[{(4+k)}^2-k^2]\ge 0$
$\Rightarrow k\ge -2$
$\Rightarrow 3a+b\ge -2$.
Ans: D