Question

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

• A. ${\left(\frac{a}{A}\right)}^2$
• B. ${\left(\frac{A}{a}\right)}^2$
• C. $0$
• D. $1$

Answer 1

The correct option is A ${\left(\frac{a}{A}\right)}^2$

$\Rightarrow$ $\alpha$ and $\beta$ are the roots of quadratic equation $a{x}^2+2bx+c=0$.

$\Rightarrow$ $\alpha \beta =\frac{c}{a}$ ----- ( 1 )

$\Rightarrow$ $\alpha +\beta =\frac{-2b}{a}$

$\Rightarrow$ $(\alpha +\beta {)}^2={\alpha }^2+{\beta }^2+2\alpha \beta$

$\Rightarrow$ ${\left(\frac{-2b}{a}\right)}^2={\alpha }^2+{\beta }^2+2\times \frac{c}{a}$

$\therefore$ ${\alpha }^2+{\beta }^2=\frac{4b^2}{a^2}-\frac{2c}{a}$ ----- ( 2 )

$\Rightarrow$ $(\alpha -\beta {)}^2={\alpha }^2+{\beta }^2-2\alpha \beta$

$\Rightarrow$ $(\alpha -\beta {)}^2=\frac{4b^2}{a^2}-\frac{2c}{a}-\frac{2c}{a}$ [ From ( 1 ) and ( 2 ) ]

$\Rightarrow$ $(\alpha -\beta {)}^2=\frac{4b^2-4ac}{a^2}$

$\therefore$ $\alpha -\beta =\sqrt{\frac{4b^2-4ac}{a^2}}$ ------ ( 3 )

$\Rightarrow$ Now, $\alpha +\delta$ and $\beta +\delta$ are roots of quadratic equation $A{x}^2+2Bx+C=0$

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

$\Rightarrow$ $\alpha \beta +\alpha \delta +\beta \delta +{\delta }^2=\frac{C}{A}$

$\therefore$ $\alpha \beta =\frac{C}{A}-\alpha \delta -\beta \delta -{\delta }^2$ -------- ( 4 )

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

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

$\Rightarrow$ $\alpha +\beta =\frac{-2B}{A}-2\delta$ -------- ( 5 )

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

$\therefore$ ${\alpha }^2+{\beta }^2={(\frac{-2B}{A}-2\delta )}^2-2\alpha \beta$ -------- ( 6 )

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

$\Rightarrow$ $(\alpha -\beta {)}^2={\alpha }^2+{\beta }^2-2\alpha \beta$

$\Rightarrow$ $(\alpha -\beta {)}^2={(\frac{-2B}{A}-2\delta )}^2-2\alpha \beta -2\alpha \beta$ [ From ( 6 ) ]

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

$\Rightarrow$ $(\alpha -\beta {)}^2=\frac{4B^2}{A^2}+\frac{8B\delta }{A}+4{\delta }^2-\frac{4C}{A}+4\alpha \delta +4\beta \delta +4{\delta }^2$ [ From ( 4 ) ]

$\Rightarrow$ $(\alpha -\beta {)}^2=\frac{4B}{A^2}-\frac{4C}{A}+\frac{8B\delta }{A}+8{\delta }^2+4\delta (\alpha +\beta )$

$\Rightarrow$ $(\alpha -\beta {)}^2=\frac{4B^2-4AC}{A^2}+\frac{8B\delta }{A}+8{\delta }^2-\frac{8B\delta }{A}-8{\delta }^2$ [ From ( 5 ) ]

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

$\therefore$ $(\alpha -\beta )=\sqrt{\frac{4B^2-4AC}{A^2}}$ ------- ( 7 )

$\Rightarrow$ $\frac{\alpha -\beta }{\alpha -\beta }=\frac{\sqrt{\frac{4b^2-4ac}{a^2}}}{\sqrt{\frac{4B^2-4AC}{A^2}}}$ [ Dividing ( 3 ) by ( 7 )]

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

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

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