Question

If $\alpha ,\beta$ be the roots of $a{x}^2+2bx+c=0$ and $\alpha +\delta ,\beta +\delta$ be those of $A{x}^2+2Bx+C=0$ for some constant $\delta$, then prove that $\frac{b^2-ac}{B^2-AC}={\left(\frac{a}{A}\right)}^2$

Answer 1

$a{x}^2+2bx+c=0$ 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 $\alpha +\delta ,\beta +\delta$

$\therefore \alpha +\delta +\beta +\delta =\frac{-2B}{A}$

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

$\therefore {\left[\right(\alpha +\delta )-(\beta +\delta \left)\right]}^2=\left[\right(\alpha +\delta )+(\beta +\delta ){]}^2=4(\alpha +\delta \left)\right(\beta +\delta )$

$\Rightarrow (\alpha -\beta {)}^2=\frac{4B^2}{A^2}-\frac{4C}{A}=\frac{4}{A^2}(B^2-AC)$

$\therefore \frac{\frac{4b^2-4ac}{a^2}}{\frac{4B^2-4AC}{A^2}}=1$

$\Rightarrow \frac{b^2-ac}{B^2-AC}=\frac{a^2}{A^2}={\left(\frac{a}{A}\right)}^2$ (proved)