Question

If $a+b+c>\frac{9c}{4}$ and quadratic equation $a{x}^2+2bx-5c=0$ has non-real, then

• A. $a>0,c>0$
• B. $a>0,c<0$
• C. $a<0,c<0$
• D. $a<0,c>0$

Answer 1

The correct option is B $a>0,c<0$

$a+b+c>\frac{9c}{4}$ $\&$ $4b^2+20ac<0$

$\Rightarrow a+b>\frac{5c}{4}$ $b^2+5ac<0$

$\Rightarrow b>\frac{5c}{4}-a$

$\Rightarrow a>0\&c<0.$

Hence, the answer is $a>0,c<0.$