Question

If $a\ne 0$ and equation $a{x}^2+bx+c=0$ has two roots $\alpha$ and $\beta$ such that $\alpha <-3$ and $\beta >2,$ which of the following is always true ?

• A. $a(a+|b|+c)>0$
• B. $a(a+|b|+c)<0$
• C. $9a-3b+c>0$
• D. $(9a-3b+c)(4a+2b+c)<0$

Answer 1

The correct option is B $a(a+|b|+c)<0$

$\because \alpha <-3<2<\beta$
$\Rightarrow -3,2$ lies between roots.
$a{x}^2+bx+c=0$
$\Rightarrow x^2+\frac{b}{a}x+\frac{c}{a}=0$

$f(-3)<0$
$\Rightarrow 9-\frac{3b}{a}+\frac{c}{a}<0$
$\Rightarrow a(9a-3b+c)<0$

$f\left(2\right)<0$
$\Rightarrow 4+\frac{2b}{a}+\frac{c}{a}<0$
$\Rightarrow a(4a+2b+c)<0$

Also,
$f(-1)<0\Rightarrow a(a-b+c)<0$
$f\left(1\right)<0\Rightarrow a(a+b+c)<0$
$\Rightarrow a(a+|b|+c)<0$