Question

If both the roots of $a{x}^2+bx+c=0$ are real, positive and distinct, then
(where $\Delta =b^2-4ax$)

• A. $\Delta >0,ab<0,ac>0$
• B. $\Delta >0,ab<0,ac<0$
• C. $\Delta >0,ab>0,bc>0$
• D. $\Delta >0,ab<0,bc>0$

Answer 1

The correct option is A $\Delta >0,ab<0,ac>0$

$\because$ Roots are real, positive and distinct.

$\therefore \Delta >0$ and graph of $f\left(x\right)=a{x}^2+bx+c$ will cut the $x-$axis in positive direction.

Case $1:$ $a<0$

Here, $c<0$ because curve intersects with $y-$ axis in $-ve$ region
$-\frac{b}{2a}>0\Rightarrow b>0(\because x\text{coordinate of vertex is}+ve$
$\Rightarrow ab<0,ac>0$

Case $2:a>0$

Here, $c>0$ because curve intersects with $y-$ axis in $+ve$ region.
$-\frac{b}{2a}>0\Rightarrow b<0(\because x\text{coordinate of vertex is}+ve$
$\Rightarrow ab<0,ac>0$

Alternate solution:
For real, positive and distinct roots.
$\Delta >0$, sum of roots $>0$, product of root $>0$
$\Rightarrow \Delta >0,-\frac{b}{a}>0,\frac{c}{a}>0\Rightarrow \Delta >0,-\frac{ab}{a^2}>0,\frac{ac}{a^2}>0\Rightarrow \Delta >0,ab<0,ac>0$