Question

If the roots of the quadratic equation $a(x-1{)}^2+2b(x-2)+c(x-1)+4=0$ are imaginary, where $a,b,c\in R$ and $b>2$, then

• A. $a+c<-4$
• B. $a+c>-4$
• C. $3c+8b-9a>4$
• D. $3c+8b-9a<4$

Answer 1

Answer: (A), (C)

$3c+8b-9a>4$

Let, $f\left(x\right)=a(x-1{)}^2+2b(x-2)+c(x-1)+4$
$f\left(1\right)=-2b+4\Rightarrow f\left(1\right)<0(\because b>2)$
$\therefore f\left(x\right)<0\forall x\in R$

$f\left(2\right)=a+c+4<0\Rightarrow a+c<-4$
$f(-2)=9a-8b-3c+4<0$
$\Rightarrow 3c-9a+8b>4$