Question

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

• A. 3
• B. -3
• C. 2
• D. -2
  

Answer 1

The correct option is D -2



Solution : $f\left(x\right)=a{x}^2+bx+6=0$
f(0) = 6 positive
$f\left(x\right)=a{x}^2+bx+6\ge 0\forall x\in R$
$f\left(3\right)=9a+3b+6\ge 0$
$3(3a+b)\ge -6$
$3a+b\ge -2$