Question

If $\alpha ,\beta$ are the roots of $a{x}^2+2bx+c=0$ and $a+\delta ,\beta +\delta$ are the roots of $A{x}^2+2Bx+C=0$ then $\frac{b^2–ac}{B^2-AC}$ is

• A. $\frac{A}{a}$
• B. $\frac{a}{A}$
• C. ${\left(\frac{A}{a}\right)}^2$
• D. ${\left(\frac{a}{A}\right)}^2$

Answer 1

The correct option is D

${\left(\frac{a}{A}\right)}^2$

$a{x}^2+2bx+c=0$ have roots $\alpha \&\beta$
$\therefore \alpha +\beta =\frac{-2b}{a},\alpha \beta =\frac{c}{a}$
$(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta$
$\Rightarrow (\alpha -\beta {)}^2=\frac{4b^2}{a^2}-\frac{4c}{a}=\frac{4}{a^2}(b^2-ac)$

$\because A{x}^2+2Bx+C=0$ have roots $\alpha +\delta \&\beta +\delta$

$\alpha +\delta +\beta +\delta =\frac{-2B}{A}$
$(\alpha +\delta )(\beta +\delta )=\frac{C}{A}$

${\left[\right(\alpha +\delta )-(\beta +\delta \left)\right]}^2={\left[\right(\alpha +\delta )+(\beta +\delta \left)\right]}^2-4(\alpha +\delta )(\beta +\delta )$
$\Rightarrow (\alpha -\beta {)}^2=\frac{4B^2}{A^2}-\frac{4C}{A}=\frac{4}{A^2}(B^2-AC)$
$\Rightarrow \frac{\frac{4}{a^2}(b^2-ac)}{\frac{4}{A^2}(B^2-AC)}$
$\Rightarrow \frac{b^2-ac}{B^2-AC}=\frac{a^2}{A^2}={\left(\frac{a}{A}\right)}^2$