Question

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

• A. $\frac{a+b+c}{A+B+C}$
• B. $\frac{b^2}{B^2}$
• C. $\frac{a^2}{A^2}$
• D. $\frac{c^2}{C^2}$

Answer 1

The correct option is C $\frac{a^2}{A^2}$

Given: $\alpha ,beta$ are the roots of the equation $a{x}^2+2bx+c=0$
$\Rightarrow \alpha +\beta =-\frac{2b}{a}$
$\&\alpha \beta =\frac{c}{a}$
Now, Difference of Roots
$(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta \Rightarrow (\alpha -\beta {)}^2=\frac{4b^2}{a^2}-4\frac{c}{a}\Rightarrow (\alpha -\beta {)}^2=\frac{4(b^2-ac)}{a^2}\cdots \left(i\right)$
Similarly, $\alpha +h,\beta +h$ are the roots of $A{x}^2+2Bx+C=0$
$\therefore$ Difference of Roots $\left[\right(\alpha +h)-(\beta +h){]}^2=[(\alpha +h)+(\beta +h){]}^2-4(\alpha +h)(\beta +h)\Rightarrow (\alpha -\beta {)}^2=\frac{4B^2}{A^2}-4\frac{C}{A}\Rightarrow (\alpha -\beta {)}^2=\frac{4(B^2-AC)}{A^2}\cdots \left(ii\right)$
Thus, comparing $\left(i\right)\&\left(ii\right),$ we get:
$\frac{4(b^2-ac)}{a^2}=\frac{4(B^2-AC)}{A^2}\Rightarrow \frac{b^2-ac}{B^2-AC}=\frac{a^2}{A^2}$