Question

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

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

Answer 1

Answer: (A)

$<0$

$\Rightarrow$ Let $f\left(x\right)=a{x}^2+2bx-3c$

$\Rightarrow$ Since $f\left(x\right)=0$ has non real roots, $f\left(x\right)$ will have the same sign for values of $x$.

It is given, $\frac{3c}{4}<a+b\Rightarrow 4a+4b-3c>0$

$\Rightarrow$ $f\left(2\right)=a(2{)}^2+2b(2)-3c=4a+4b-3c>0$ [ From above ]

and $f\left(0\right)=-3c>0$

$\Rightarrow$ $c<0$