Question

If $a{x}^2+2bx+3c=0,a\ne 0,c>0$ does not have any real roots, then which of the following is/are true?

• A. $4a-4b+3c>0$
• B. $a+6b+27c>0$
• C. $4a-4b+3c<0$
• D. $a+6b+27c<0$

Answer 1

Answer: (A), (B)

The correct options are
A $4a-4b+3c>0$
B $a+6b+27c>0$

Let $f\left(x\right)=a{x}^2+2bx+3c$
Now, $f\left(x\right)=0$ does not have real root,
Checking $f\left(0\right)=3c>0$, so
$f\left(x\right)>0\forall x\in R$

Now,
$f(-2)>0\Rightarrow 4a-4b+3c>0$

$f\left(\frac{1}{3}\right)>0\Rightarrow \frac{a}{9}+\frac{2b}{3}+3c>0\Rightarrow a+6b+27c>0$