Question

If the equation $a{x}^2+2bx-3c=0$ has non real roots and $\left(3c/4\right)<(a+b);$ then c is always

• A. < 0
• B. > 0
• C. $\ge 0$
• D. zero

Answer 1

The correct option is A < 0

Let $f\left(x\right)=a{x}^2+2bx-3c$
Since, f(x) = 0 has non real roots, f(x) will have the same sign for all values of x
Given $\frac{3c}{4}<a+b$$\Rightarrow$ 4a + 4b - 3c > 0
$\Rightarrow f\left(2\right)>0\Rightarrow f\left(0\right)>0\Rightarrow c<0.$