Question

If $2a{x}^2+3bx+5c=0,a\in R-\left\{0\right\},c>0$ does not have any real roots, then which of the following is/are always true?

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

Answer 1

Answer: (A), (D)

The correct options are
A $a>0$
D $4a-3b+3c>0$

Given $2a{x}^2+3bx+5c=0,a\in R-\left\{0\right\},c>0$ does not have any real roots.

Let $f\left(x\right)=2a{x}^2+3bx+5c$
$\Rightarrow f\left(x\right)>0,\forall x\in R\text{if}a>0\Rightarrow f\left(x\right)<0,\forall x\in R\text{if}a<0$

Putting $x=0$, we get
$f\left(0\right)=5c\Rightarrow f\left(0\right)>0(\because c>0)\therefore f\left(x\right)>0,\forall x\in R$
So $a>0$
Putting $x=-2$
$\Rightarrow f(-2)>0\Rightarrow 8a-6b+5c>0\Rightarrow 8a-6b+6c>c>0\Rightarrow 4a-3b+3c>0$